学步园

先把多源多汇网络转化为单源单汇网络，然后EK模版

code

#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define lson l,m,rt << 1
#define rson m+1,r,rt<<1|1
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("input.txt","r",stdin)
#define writef freopen("output.txt","w",stdout)
const int maxn = 102;
const int INF = 0x3fffffff;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};
const double pi = acos(-1.0);
const double eps= 1e-7;
using namespace std;
int n,np,nc,m;
int cap[maxn][maxn],pre[maxn];
bool vis[maxn];
bool bfs(int s,int t){
    int p;
    queue<int> q;
    SET(pre,-1);SET(vis,false);
    pre[s]=-1;
    q.push(s);
    while(!q.empty()){
        p=q.front();q.pop();
        if(p==n+1) return true;
        FOR(i,0,t){
            if(cap[p][i] && !vis[i]){
                pre[i]=p;
                vis[i]=true;
                q.push(i);
            }
        }
    }
    return false;
}
int EK(int s,int t){
    int flow=0,d;
    while(bfs(s,t)){
        d=INF;
        for(int i=t;i!=s;i=pre[i])
            d=min(cap[pre[i]][i],d);
        for(int i=t;i!=s;i=pre[i]){
            cap[pre[i]][i]-=d;
            cap[i][pre[i]]+=d;
        }
        flow+=d;
    }
    return flow;
}
int main()
{
    int a,b,z;
    while(~scanf("%d%d%d%d",&n,&np,&nc,&m)){
        SET(cap,0);
        FOR(i,1,m){
            scanf(" (%d,%d)%d",&a,&b,&z);
            cap[a+1][b+1]=z;
        }
        FOR(i,1,np){
            scanf(" (%d)%d",&a,&z);
            cap[0][a+1]=z;
        }
        FOR(i,1,nc){
            scanf(" (%d)%d",&a,&z);
            cap[a+1][n+1]=z;
        }
        PD(EK(0,n+1));LN;
    }
    return 0;
}