愉快的校赛翻皮水！

题解

A 温暖的签到，注意用gets

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn =  + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

string a;

int main(){

    while(getline(cin,a)){

        a.back() = '!';

        cout << a << endl;

    }

    return ;

}

A

B.比赛的时候一直以为是主席树上操作或者其他的高级数据结构,万万没想到是在序列特性下手,打一张最小的不冲突的表就会发现斐波那契数列是最小不冲突序列,int范围内最多容纳47个数左右,所以小于50的范围暴力查询,大于50的范围必定YES

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

LL a[maxn];

LL b[maxn];

int main(){

    Sca2(N,M);

    for(int i = ; i <= N ; i ++) Scl(a[i]);

    for(int i = ; i <= M ; i ++){

        int l,r; Sca2(l,r);

        if(r - l +  < ){

            puts("NO");

            continue;

        }

        if(r - l +  >= ) puts("YES");

        else{

            int flag = ;

            for(int j = l; j <= r; j ++) b[j] = a[j];

            sort(b + l,b + r + );

            for(int j = l + ; j <= r; j ++){

                if(b[j - ] + b[j - ] > b[j]){

                    flag = ;

                    break;

                }

            }

            if(flag) puts("YES");

            else puts("NO");

        }

    }

    return ; //1 1 2 3 5 8 13 21 34

}

B

D. 8说了,温暖的签到

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int a[maxn];

set<int>Q;

int main(){

    N = read();

    while(N--) Q.insert(read());

    Pri(Q.size());

    return ; //1 1 2 3 5 8 13 21 34

}

D

E.给一个条件构造的图，求图上的哈密顿回路。

可以猜想到N为奇数的时候始终不可行。

N为偶数的时候必定可行，并且可以发现，每个点都恰好有两个出度和两个入度，

这就可以转换成欧拉回路直接做

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int ans[maxn],cnt;

bool vis[maxn];

void dfs(int t){

    int l = (t * ) % N,r = (t *  + ) % N;

    if(l > r) swap(l,r);

    if(!vis[r]){vis[r] = ;dfs(r);}

    if(!vis[l]){vis[l] = ;dfs(l);}

    ans[++cnt] = t;

}

int main(){

    Sca(N);

    if(N & ){puts("-1"); return ;}

    dfs();

    for(int i = cnt ; i >= ; i --)cout << ans[i] << " " ;

    return ;

}

E

G.由于每天加的钱为实数而不要求为浮点数,一个显然的贪心是每两个取的物品i,j之间,每天加入的钱都是wj / (j - i)

dp是显然的,第一个难点在于每天钱数不增的限制,如果dp存储加入的钱数,2000 * 1e6很显然时间复杂度上过不去

一个比较巧妙地思想是dp[i][j]表示上一个操作是i - > j的物品的选择,2000 * 2000满足了时间复杂度还满足了钱数

得到状态转移方程dp[j][k] = max(dp[i][j] + V[k])

到了这一步就可以写出一个n3 的暴力(雾),将状态转移方程变形,得到i > j - W(j) / W(k) * (k - j)这样一个右边和i无关的方程.

所以考虑枚举j和k,然后就可以得到i的下界,i的上界显然为j - 1,就变成了一个后缀最大值的问题,这个甚至不需要树状数组,直接维护一个简单的一维数组即可.

时间复杂度n²

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N;

struct Good{

    LL v;

    double w;

}P[maxn];

LL dp[maxn][maxn];

LL Max[maxn];

int main(){

    Sca(N);

    for(int i = ; i <= N ; i ++) scanf("%lf",&P[i].w);

    for(int i = ; i <= N ; i ++) Scl(P[i].v);

    for(int i = ; i <= N ; i ++){

        for(int j = ; j <= N ; j ++){

            dp[i][j] = -1e18;

        }

    }

    for(int i = ; i <= N ; i ++) dp[][i] = P[i].v;

    for(int j = ; j <= N ; j ++){

        Max[j] = -1e18;

        for(int k = j - ; k >= ; k --) Max[k] = max(Max[k + ],dp[k][j]);

        for(int k = j + ; k <= N ; k ++){

            double l;

            if(!P[k].w) l = ;

            else l = max(j - P[j].w / P[k].w * (k - j),(double));

            int L = (int)l;

            if(fabs(l - L) > eps) L++;

            if(L > j - ) continue;

            dp[j][k] = Max[L] + P[k].v; // Max[k][j]

        }

    }

    LL ans = ;

    for(int i = ; i <= N ; i ++){

        for(int j = i + ; j <= N ; j ++){

            ans = max(ans,dp[i][j]);

        }

    }

    Prl(ans);

    return ;

}

G

H.取个log就会发现变成AjlogAi > AilogAj,直接sort一波,特判一下前后相等的情况即可.

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 1e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

PIL a[maxn];

int ans[maxn];

bool cmp(PIL x,PIL y){

    return 1.0 * x.se * log(y.se * 1.0) > 1.0 * y.se * log(x.se * 1.0);

}

bool equal(double x,double y){

    return fabs(x - y) < eps;

}

int main(){

    N = read();

    for(int i = ; i <= N ; i ++){

        Scl(a[i].se);

        a[i].fi = i;

    }

    sort(a + ,a +  + N,cmp);

    int cnt = ;

    for(int i = ; i <= N ; i ++){

        if(i >  && equal(1.0 * a[i].se * log(a[i - ].se * 1.0),1.0 * a[i - ].se * log(a[i].se * 1.0))) cnt++;

        else cnt = ;

        ans[a[i].fi] = i -  - cnt;

    }

    for(int i = ; i <= N ; i ++){

        printf("%d ",ans[i]);

    }

    return ;

}

H

I.要有多坑有多坑,正确题意为双射 + 最后25字母可推26字母,其他没有难度

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 1e6 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int to[],to2[];

char str[maxn],str2[maxn];

int main(){

    while(~scanf("%s%s",str,str2)){

        int l = strlen(str);

        for(int i =  ; i < ; i ++) to[i] = to2[i] = -;

        for(int i = ; str[i]; i ++){

            int id1 = str[i] - 'a',id2 = str2[i] - 'a';

            if(~to[id1] && to[id1] != id2){

                puts("Impossible"); exit();

            }

            if(~to2[id2] && to2[id2] != id1){

                puts("Impossible"); exit();

            }

            to[id1] = id2; to2[id2] = id1;

        }

        int cnt = ;

        for(int i =  ; i < ; i ++) if(~to[i]) cnt++;

        if(cnt == ){

            int t = ;

            for(int i =  ; i < ; i ++){

                if(to[i] == -) t = i;

            }

            for(int i =  ; i < ; i ++) if(to2[i] == -) to[t] = i;

        }

        for(int i =  ; i < ; i ++){

            if(~to[i]) printf("%c->%c\n",i + 'a',to[i] + 'a');

        }

    }

    return ;

}

I

J.过于温暖,给出题人点赞

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

LL N,M,K,T;

int main(){

    cin >> N >> K >> T;

    cout << max(0LL,N - K * T) << endl;

    return ;

}

J

K.很显然是预处理出所有情况然后二分端点.

坑点1.不能用海伦公式,过不了double浮点数的误差,需要用向量叉积并且不除2,后面查询的时候将l和r乘2达到无浮点数的目的

2.r用upper_bound,l用lower_bound,最后r - l即可,天知道为什么我一开始手写了两个二分

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,Q;

struct Point{

    LL x,y;

}point[maxn];

LL P[ *  * ];

LL cul(LL x1,LL y1,LL x2,LL y2){

    //cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl;

    return abs(x1 * y2 - x2 * y1);

}

int main(){

    Sca2(N,Q);  //x1:(ax - bx) y1:(ay - by) x2:ax - cx y2:ay - cy

    for(int i = ; i <= N ; i ++) scanf("%lld%lld",&point[i].x,&point[i].y);

    int cnt = ;

    for(int i = ; i <= N ; i ++){

        for(int j = i + ; j <= N; j ++){

            LL x1 = point[i].x - point[j].x,y1 = point[i].y - point[j].y;

            for(int k = j + ; k <= N ; k ++){

                LL x2 = point[i].x - point[k].x,y2 = point[i].y - point[k].y;

                P[++cnt] = cul(x1,y1,x2,y2);

            }

        }

    }

    sort(P + ,P +  + cnt);

    //for(int i = 1; i <= cnt; i ++) cout << P[i] << " ";

    //cout << endl;

    for(int i = ; i <= Q; i ++){

        LL l,r; scanf("%lld%lld",&l,&r);

        l *= ; r *= ;

        l = lower_bound(P + ,P +  + cnt,l) - P;

        r = upper_bound(P + ,P +  + cnt,r) - P;

        //cout << l << ' ' << r << endl;

        Pri(r - l);

    }

    return ;

}

K

L.考虑从T开始跑一颗最短路树，那么如果当人在i点的时候去掉边，显然去掉非树边上的边的行为是无意义的，如果去掉了树边上的边，那么就需要走到他们的子树上某个结点j，通过一个非树边走到一个非子树的结点k，然后再到T点，距离就是dis[j] - dis[i] + edge(j,k) + dis[k]

显然（雾）我们可以对每一条非树边进行预处理，每一条连接(i,j)的非树边将会对i到lca(i,j),j到lca(i,j)的两条链产生贡献（目的是寻找一条i点树边被封之后最短的走到T的路线),这个过程可以用树链剖分实现

最后再从T开始往S跑最短路，由于我们已经求出了每个点i的树边被封之后到T的最短路，所以更新答案是dis[v] = max(dis[u.pos] + edge[i].dis,dis2[v]);

也就是说，如果在v点直接切断会比在之后切断造成需要走更长的路。

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 2e5 + ;

const int maxm = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,S,T;

struct Edge{

    int to,next;

    LL dis;

}edge[maxm * ];

struct E{

    int u,v,flag;

    LL w;

    E(){}

    E(int u,int v,LL w):u(u),v(v),w(w){}

}e[maxm * ];

int head[maxn],tot;

void init(){

    for(int i =  ; i <= N ; i ++) head[i] = -;

    tot = ;

}

void add(int u,int v,LL w){

    edge[tot].to = v;

    edge[tot].next = head[u];

    edge[tot].dis = w;

    head[u] = tot++;

}

struct node{

    int pos;

    LL cost;

    node(){}

    node(int pos,LL cost):pos(pos),cost(cost){}

    friend bool operator < (node a,node b){

        return a.cost > b.cost;

    }

};

LL dis[maxn],dis2[maxn];

int fa[maxn];

void Dijkstra(int s){

    for(int i = ; i <= N ; i ++) dis[i] = 1e18;

    priority_queue<node>Q;

    Q.push(node(s,)); fa[s] = dis[s] = ;

    while(!Q.empty()){

        node u = Q.top(); Q.pop();

        if(u.cost > dis[u.pos]) continue;

        for(int i = head[u.pos]; ~i; i = edge[i].next){

            int v = edge[i].to;

            if(dis[v] > dis[u.pos] + edge[i].dis){

                dis[v] = dis[u.pos] + edge[i].dis;

                fa[v] = u.pos;

                Q.push(node(v,dis[v]));

            }

        }

    }

}

int dep[maxn],top[maxn],pos[maxn],size[maxn],son[maxn];

void dfs1(int t,int la){

    size[t] = ; son[t] = t;

    int heavy = ;

    for(int i = head[t]; ~i ; i = edge[i].next){

        int v = edge[i].to;

        if(v == la) continue;

        dep[v] = dep[t] + ;

        fa[v] = t;

        dfs1(v,t);

        if(size[v] > heavy){

            heavy = size[v];

            son[t] = v;

        }

        size[t] += size[v];

    }

}

int cnt;

void dfs2(int t,int la){

    top[t] = la;

    pos[t] = ++cnt;

    if(son[t] == t) return;

    dfs2(son[t],la);

    for(int i = head[t]; ~i ; i = edge[i].next){

        int v = edge[i].to;

        if((fa[t] == v) || v == son[t]) continue;

        dfs2(v,v);

    }

}

int lca(int u,int v){

    while(top[u] != top[v]){

        if(dep[top[u]] < dep[top[v]]) swap(u,v);

        u = fa[top[u]];

    }

    if(dep[u] < dep[v]) return u;

    return v;

}

struct Tree{

    int l,r;

    LL lazy;

}tree[maxn << ];

void Build(int t,int l,int r){

    tree[t].l = l; tree[t].r = r;

    tree[t].lazy = 1e18;

    if(l == r) return;

    int m = l + r >> ;

    Build(t << ,l,m); Build(t <<  | ,m + ,r);

}

void Pushdown(int t){

    tree[t << ].lazy = min(tree[t << ].lazy,tree[t].lazy);

    tree[t <<  | ].lazy = min(tree[t <<  | ].lazy,tree[t].lazy);

}

void update(int t,int l,int r,LL sum){

    if(l <= tree[t].l && tree[t].r <= r){

        tree[t].lazy = min(tree[t].lazy,sum);

        return ;

    }

    Pushdown(t);

    int m = tree[t].l + tree[t].r >> ;

    if(r <= m) update(t << ,l,r,sum);

    else if(l > m) update(t <<  | ,l,r,sum);

    else{

        update(t << ,l,m,sum);

        update(t <<  | ,m + ,r,sum);

    }

}

LL query(int t,int p){

    if(tree[t].l == tree[t].r) return tree[t].lazy;

    Pushdown(t);

    int m = tree[t].l + tree[t].r >> ;

    if(p <= m) return query(t << ,p);

    else return query(t <<  | ,p);

}

void update(int u,int v,LL sum){

    while(top[u] != top[v]){

        update(,pos[top[u]],pos[u],sum);

        u = fa[top[u]];

    }

    if(u == v) return;

    update(,pos[v] + ,pos[u],sum);

}

int main(){

    scanf("%d%d%d%d",&N,&M,&S,&T); init();

    for(int i = ; i <= M ; i ++){

        int u,v; Sca2(u,v); LL w; Scl(w);

        add(u,v,w); add(v,u,w);

        e[i] = E(u,v,w); e[i].flag = ;

    }

    Dijkstra(T); init();

    for(int i = ; i <= M ; i ++){

        if(fa[e[i].u] == e[i].v || fa[e[i].v] == e[i].u){

            add(e[i].v,e[i].u,e[i].w);

            add(e[i].u,e[i].v,e[i].w);

        }else e[i].flag = ;

    }

    dfs1(T,T); dfs2(T,T);

    Build(,,N);

    for(int i = ; i <= M ; i ++){

        if(e[i].flag) continue;

        LL sum = dis[e[i].u] + dis[e[i].v] + e[i].w;

        int l = lca(e[i].u,e[i].v);

        update(e[i].u,l,sum); update(e[i].v,l,sum);

    }

    for(int i = ; i <= N; i ++) dis2[i] = query(,pos[i]) - dis[i];

    for(int i = ; i <= N ; i ++) dis[i] = 1e18;  init();

    for(int i = ; i <= M ; i ++){

        add(e[i].u,e[i].v,e[i].w);

        add(e[i].v,e[i].u,e[i].w);

    }

    priority_queue<node>Q;

    Q.push(node(T,)); dis[T] = ;

    while(!Q.empty()){

        node u = Q.top(); Q.pop();

        if(dis[u.pos] < u.cost) continue;

        for(int i = head[u.pos]; ~i ; i = edge[i].next){

            int v = edge[i].to;

            if(dis[v] > max(dis[u.pos] + edge[i].dis,dis2[v])){

                dis[v] = max(dis[u.pos] + edge[i].dis,dis2[v]);

                Q.push(node(v,dis[v]));

            }

        }

    }

    if(dis[S] >= 1e16) puts("-1");

    else Prl(dis[S]);

    return ;

}

L

M.用SG函数打表博弈，反正我是不会做，贴上队友代码

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <stdlib.h>

#include <map>

#include <set>

#include <iostream>

#include <string>

#include <algorithm>

#include <vector>

#include <queue>

using namespace std;

typedef long long LL;

const int maxn = 1e6+;

int if_pre[maxn +];

void pre_first()

{

    memset(if_pre, 0x3f, sizeof(if_pre));

    if_pre[] = ;

    if_pre[] = ;

    int pre_num =;

    for (int i = ; i < maxn; i++)

    {

        if (if_pre[i] == 0x3f3f3f3f)

        {

            if_pre[i] = pre_num++;

            for (int j = i << ; j < maxn; j += i)

            {

                if_pre[j] = min(if_pre[j],if_pre[i]);

            }

        }

    }

}

int main()

{

    int t;

    cin >>t;

    pre_first();

    while (t --)

    {

        int n,a;

        cin >>n;

        int sum =;

        while (n--)

        {

            cin >>a;

            sum^=if_pre[a];

            //cout<<a<<'*'<<if_pre[a]<<endl;

        }

        if (sum)

            cout<<"Subconscious is our king!"<<endl;

        else

            cout<<"Long live with King Johann!"<<endl;

    }

#ifdef VSCode

    system("pause");

#endif

    return ;

}

M

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