A.维护一个前缀最大值，不断跳即可

#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 = 1e4 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int a[maxn];

int Max[maxn];

int main(){

    Sca(N);

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

        Sca(a[i]);

        a[i] = max(a[i],a[i - ]);

    }

    int cnt = ;

    int now = ;

    while(now < N){

        cnt++;

        now++;

        while(now != a[now]) now = a[now];

    }

    Pri(cnt);

    return ;

}

A

B.从左边开始删除一直删到>或者从右边开始删除一直删到<，两种方案取最小值。

#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;

char str[maxn];

int main(){

    int T = read();

    while(T--){

        Sca(N);

        scanf("%s",str + );

        int ans = N - ;

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

            if(str[i] == '>'){

                ans = min(ans,i - );

                break;

            }

        }

        for(int i = N ; i >= ; i --){

            if(str[i] == '<'){

                ans = min(ans,N - i);

                break;

            }

        }

        Pri(ans);

    }

    return ;

}

B

C.按照beauty值从大到小排序，维护一个前缀和（最多K个length）,小根堆维护一下每次要踢出哪个最小lenght的弟弟

#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 = 3e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

PLL P[maxn];

bool cmp(PLL a,PLL b){

    return a.se > b.se;

}

int main(){

    Sca2(N,K);

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

        scanf("%lld%lld",&P[i].fi,&P[i].se);

    }

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

    priority_queue<LL,vector<LL>,greater<LL>>Q;

    LL sum = ;

    LL ans = ;

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

        ans = max(ans,(sum + P[i].fi) * P[i].se);

        Q.push(P[i].fi); sum += P[i].fi;

        if(Q.size() >= K){

            sum -= Q.top();

            Q.pop();

        }

    }

    Prl(ans);

    return ;

}

C

D.发现1,2,3| 1,3,4| 1,4,5 .. |1,n - 1,n为最优解。

#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 = 3e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int main(){

    Sca(N);

    LL sum = ;

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

        sum += i * (i + );

    }

    Prl(sum);

    return ;

}

D

E*

1.发现不存在奇长度的回文串只需要满足不存在长度3的回文串即可，长度为5的回文串包含了长度3的回文串。

2.也就是说，要满足所有的str[x] != str[x + 2],发现奇偶位置上的数字其实互不干扰，那可以把序列分成两个序列来做。

3.得到dp方程,dp[i][j]表示到了i这个位置，且这个位置上的数字为j的时候，满足条件序列的总数，状态转移方程为dp[i][j] = ∑dp[i - 1][p] (1 <= p <= K && p != j)

4.当然这个时间复杂度和空间复杂度双双nk的算法是行不通的，我们发现对于i相同的dp[i][j],只会有两种不同的值，并且是其中k - 1个值相同以及另一个值鹤立鸡群的情况（当然也有可能都是鸡完全相同）

5.考虑标注特殊的点位置和特殊的点值以及k - 1个其他的点值，即可压缩复杂度为O(n)

#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 = ;

int N,M,K;

LL a[maxn],b[maxn];

LL solve(LL *x,int n){

    LL big,small,pos;

    if(x[] == -){

        big = ; small = ; pos = ;

    }else{

        big = ; small = ; pos = x[];

    }

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

        if(x[i] == -){

            LL s = big * (K - ),b = big * (K - )  + small;

            small = s % mod; big = b % mod;

        }else{

            if(pos == x[i]){

                small = big * (K - ) % mod;

                big = ;

            }else{

                small = big * (K - ) + small; small %= mod;

                pos = x[i];

                big = ;

            }

        }

    }

    return (small + big * (K - )) % mod;

}

int main(){

    Sca2(N,K);

    int cnt1 = ,cnt2 = ;

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

        if(i & ) a[++cnt1] = read();

        else b[++cnt2] = read();

    }

    Prl(solve(a,cnt1) * solve(b,cnt2) % mod);

    return ;

}

E

F*

1.考虑将两个集合用一个二维的表来表示，添加就是将i行j列合并，最终答案是每个联通块的行数 * 列数。

2.考虑到用并查集,size1[maxn],size2[maxn],分别表示这个集合里面行列的数量。

3.问题在于删除，并查集是不存在在线直接删除这种黑科技的，但是我们可以用离线的方式，将并查集的存在生命周期放到线段树上，然后dfs整颗线段树进行优化。

这个操作在BZOJ2049里遇到过了，当时还写了blog https://www.cnblogs.com/Hugh-Locke/p/10367480.html

结果这次还是翻皮水了（雾）

#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 = 6e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

map<PII,int>P;

//并查集

int fa[maxn];

LL size1[maxn],size2[maxn];

int dep[maxn];

void init(){

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

        fa[i] = i;

        if(i > 3e5) size2[i] = ;

        else size1[i] = ;

        dep[i] = ;

    }

}

//线段树

struct Edge{

    PII to;

    int next;

}edge[maxn * ];

int tot;

void add(int& t,PII w){

    edge[tot].to = w;

    edge[tot].next = t;

    t = tot++;

}

struct Tree{

    int l,r;

    int head;

}tree[maxn << ];

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

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

    tree[t].head = -;

    if(l == r) return;

    int m = l + r >> ;

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

}

void add(int t,int l,int r,PII w){

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

        add(tree[t].head,w);

        return;

    }

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

    if(r <= m) add(t << ,l,r,w);

    else if(l > m) add(t <<  | ,l,r,w);

    else{

        add(t << ,l,m,w); add(t <<  | ,m + ,r,w);

    }

}

int Stack[maxn],top;

int find(int x){

    while(fa[x] != x) x = fa[x];

    return x;

}

LL ans;

void rewind(int t){

    while(top > t){

         int x = Stack[--top];

         dep[fa[x]] -= dep[x] + ;

         ans -= 1LL * size1[fa[x]] * size2[fa[x]];

         size1[fa[x]] -= size1[x];

         size2[fa[x]] -= size2[x];

         ans += 1LL * size1[fa[x]] * size2[fa[x]];

         ans += 1LL * size1[x] * size2[x];

         fa[x] = x;

    }

}

void dfs(int t){

    int now = top;

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

        PII t = edge[i].to;

        t.fi = find(t.fi); t.se = find(t.se);

        if(t.fi == t.se) continue;

        if(dep[t.fi] > dep[t.se]) swap(t.fi,t.se);

        Stack[top++] = t.fi;

        fa[t.fi] = t.se;

        dep[t.se] += dep[t.fi] + ;

        ans -= 1LL * size1[t.fi] * size2[t.fi]; ans -= size1[t.se] * size2[t.se];

        size1[t.se] += size1[t.fi]; size2[t.se] += size2[t.fi];

        ans += 1LL * size1[t.se] * size2[t.se];

    }

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

        printf("%lld ",ans);

    }else{

        dfs(t << );

        dfs(t <<  | );

    }

    rewind(now);

}

int main(){

    Sca(N); init();

    Build(,,N);

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

        PII p; p.fi = read(),p.se = read() + 3e5;

        if(P[p]){

            add(,P[p],i - ,p);

            P[p] = ;

        }else P[p] = i;

    }

    for(map<PII,int>::iterator it = P.begin(); it != P.end(); it++)  if((*it).se) add(,(*it).se,N,(*it).fi);

    dfs();

    return ;

}

F

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