/*Lucky_Glass*/
#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
/*以下均是模板*/
const int N = 50002;
const int M = 500005;
#define INF 0x3f3f3f3f
struct E
{int to,cap,cost,flow,next;
}e[2*M];int head[N] , ecnt;
int  pre[N];
int dis[N];
bool vis[N];
int n,m,S,T;void Clear()
{ecnt = 0;memset(head,-1,sizeof head);
}void adde(int fr,int to,int cap,int cost)
{e[ecnt]=(E){to,cap,-cost,0,head[fr]};head[fr] = ecnt++;e[ecnt]=(E){fr,0,cost,0,head[to]};head[to] = ecnt++;
}bool SPFA(int s,int t)
{memset(vis,0,sizeof vis);memset(dis,0x3f,sizeof dis);memset(pre,-1,sizeof pre);queue <int> q;q.push(s);dis[s] = 0;vis[s]=1;while (!q.empty()){int cur = q.front();q.pop();vis[cur] = false;for (int j=head[cur];j!=-1;j=e[j].next){int to = e[j].to;if (dis[to] > dis[cur] + e[j].cost && e[j].cap > e[j].flow ){dis[to] = dis[cur] + e[j].cost;pre[to] = j;if (!vis[to]){q.push(to);vis[to] = true;}}}}return pre[t] != -1;
}void MCMF (int s,int t,int &maxflow,int &mincost)
{maxflow = mincost = 0;while (SPFA(s,t)){int MIN = INF;for (int j=pre[t]; j!=-1;j=pre[e[j^1].to]){MIN = min(MIN,e[j].cap - e[j].flow);}for (int j=pre[t]; j!=-1;j=pre[e[j^1].to]){e[j].flow += MIN;e[j^1].flow -= MIN;mincost += MIN * e[j].cost;}maxflow += MIN;}
}
/*模板结束*/
#define MAXN 3000int L[MAXN+5],R[MAXN+5],Lk[MAXN+5],Kd[MAXN+5];
int main(){int TT;cin>>TT;//数据组数 while(TT--){Clear();//清空 int nn,mm,kk,ht;cin>>nn>>mm>>kk>>ht;for(int i=1;i<=mm;i++)cin>>L[i]>>R[i]>>Lk[i]>>Kd[i];S=1,T=kk+mm*2+2;
//超级源点为1，超级汇点为最后一个点（因为有kk个人节点，mm*2个电影节点，即真节点和虚拟节点，再加上一个源点） for(int i=1;i<=kk;i++)adde(S,i+1,1,0); //在人节点和源点之间连边，第i个人编号为i+1 for(int i=1;i<=kk;i++)for(int j=1;j<=mm;j++)adde(i+1,j+kk+1,1,Lk[j]); //在人和电影的真节点之间连边，第i部电影真节点编号为i+kk+1 for(int i=1;i<=mm;i++)adde(i+kk+1,i+kk+mm+1,1,0); //连接真节点和虚拟节点，第i部电影的虚拟节点编号为i+kk+mm+1 for(int i=1;i<=mm;i++)adde(i+kk+mm+1,T,1,0); //连接虚拟节点和汇点 for(int i=1;i<=mm;i++)for(int j=1;j<=mm;j++)if(i!=j&&R[i]<=L[j]){ //电影之间连边，虚拟节点连真节点 if(Kd[i]!=Kd[j])adde(i+kk+mm+1,j+kk+1,1,Lk[j]);elseadde(i+kk+mm+1,j+kk+1,1,Lk[j]-ht);}int ans1,ans2;MCMF(S,T,ans1,ans2);cout<<-ans2<<"\n"; //由于边权取了相反数，输出答案时也需要取相反数 }
}

The End

Thanks for reading!

- Lucky_Glass