A - New Generation ABC

思路：分段讨论

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

int main()

{

    ll n = read();

    if(n>=1&&n<=125) cout<<4<<endl;

    else if(n>=126&&n<=211) cout<<6<<endl;

    else cout<<8<<endl;

    return 0;

}

B - How many?

思路：暴力枚举即可。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

int main()

{

    ll s = read(), t = read();

    ll ans = 0;

    rep(i,0,s) rep(j,0,s) rep(k,0,s) if(i+j+k<=s&&i*j*k<=t) ans++;

    cout<<ans<<endl;

    return 0;

}

C - Distribution

思路：前i个的最优解，由前i-1个最优解 + \(S_{i-1}\) 过来，或者直接选当前\(T_{i}\)， 由于是个环，所以多复制一段1-n到a数组后面，遍历1-2n即可得到每个最优解。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 4e5+200;

const ll inf=0x3f3f3f3f3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll each[maxn];

ll s[maxn];

ll t[maxn];

int main()

{

    ll n = read();

    rep(i,0,2*n+1) each[i] = 1e15;

    ll pre = inf;

    rep(i,1,n) s[i] = read(); rep(i,n+1,2*n) s[i] = s[i-n];

    rep(i,1,n) t[i] = read(); rep(i,n+1,2*n) t[i] = t[i-n];

    rep(i,1,n*2)

    {

        each[i] = min(min(t[i],each[i-1]+s[i-1]),each[i]);

    }

    rep(i,n+1,n*2) cout<<each[i]<<endl;

    return 0;

}

D - Sum of Maximum Weights

思路：考虑从权值从小到大逐个每个边，使得当前边讨论时，只需要加上这条边两端点的个数相乘，再乘上权值（\(ans+= numx* numy * val\)）。这一步操作利用并查集维护即可。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

typedef struct Pos

{

    ll x;

    ll y;

    ll val;

    bool operator < (const Pos &a) const

    {

        return val < a.val;

    }

}P;

P a[maxn];

ll fa[maxn];

ll num[maxn];

ll get(ll x)

{

    return fa[x] = (fa[x]==x?x:get(fa[x]));

}

void solve()

{

    ll n = read();

    rep(i,1,n) fa[i] = i, num[i] = 1;

    rep(i,1,n-1) a[i].x = read(), a[i].y = read(), a[i].val = read();

    sort(a+1,a+1+n-1);

    ll ans = 0;

    rep(i,1,n-1)

    {

        ll x = a[i].x, y = a[i].y, val = a[i].val;

        ll fx = get(x), fy = get(y);

        if(fx != fy)

        {

            ll numx = num[fx];

            ll numy = num[fy];

            ans += numx*numy*val;

            fa[fy] = fx;

            num[fx] += num[fy];

           //cout<<i<<' '<<x<<' '<<y<<' '<<val<<' '<<numx<<' '<<numy<<' '<<val<<endl;

        }

    }

    cout<<ans<<endl;

}

int main()

{

    solve();

    return 0;

}

E - Packing Under Range Regulations

思路：先考虑暴力对每个点遍历时的策略：对于当前i，我要对所有L<=i的区间里面，贪心挑一个R最小的（R小要尽快处理掉），这样对每个i都如此维护的话，一旦发现这个挑出来的R<i，则不合法，最后检验是否每个区间都被挑完了。由于i范围很大，所以考虑优化，发现要满足L<=i，有很多点可以直接跳着走，所以当发现没有区间可以挑的时候，可以直接跳到后面最近的L上，如此往复即可。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 2e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

typedef struct Pos

{

    ll l;

    ll r;

    bool operator < (const Pos &a) const

    {

        return r > a.r;

    }

}P;

P a[maxn];

bool cmp(P a, P b)

{

    return a.l < b.l;

}

int main()

{

    int kase;

    cin>>kase;

    while(kase--)

    {

        ll n = read();

        rep(i,1,n) a[i].l = read(), a[i].r = read();

        sort(a+1,a+1+n,cmp);

        priority_queue<P,vector<P> > q;

        ll cnt = 0;

        int id = 1;

        int pos = 1;

        while(id<=n||q.size())

        {

            while(id<=n&&a[id].l<=pos) q.push(a[id]), id++;

            if(q.size()&&q.top().r<pos) break;

            if(q.size()&&q.top().r>=pos) q.pop(), cnt++;

            if(q.empty()&&id<=n) pos = a[id].l;

            else if(q.size()) pos++;

        }

        puts(cnt==n?"Yes":"No");

    }

    return 0;

}

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