想过无源汇的网络流，还是比较麻烦的，然后没往下想。。。

设s点集有一些点，

然后设s点集只有一个点

#include <cstdio>
#include <map>
#include <set>
#include <algorithm>
using namespace std;
#define N 205
struct node{
    int to, d, b, nex;
}edge[5005], edge2[5005];
int head[N], edgenum;
void add(int u, int v, int d, int b){
    node E = {v, d, b, head[u]};
    edge[edgenum] = E;
    head[u] = edgenum++;
}
int head2[N], edgenum2;
void add2(int u, int v, int d, int b){
    node E = {v, d, b, head2[u]};
    edge2[edgenum2] = E;
    head2[u] = edgenum2++;
}

void init(){memset(head, -1, sizeof head), edgenum = 0;memset(head2, -1, sizeof head2), edgenum2= 0 ;}
int n, m;
bool work(int x){
    int hehe = 0, gg = 0;
    for(int i = head[x]; ~i; i = edge[i].nex){
        hehe += edge[i].d;
    }
    for(int i = head2[x]; ~i; i = edge2[i].nex){
        gg += edge2[i].d + edge2[i].b;
    }
    return hehe>gg;
}
int main() {
    int T,u,v,a,b, Cas = 1; scanf("%d", &T);
    while(T-- > 0) {
        scanf("%d %d",&n,&m);
        init();
        while(m--) {
            scanf("%d %d %d %d",&u,&v,&a,&b);
            add(u, v, a, b);
            add2(v, u, a, b);
        }
        printf("Case #%d: ",Cas++);
        bool yes = true;
        for(int i = 1; yes && i <= n; i++) {
            if(work(i))
                yes = false;
        }
        yes ? puts("happy"):puts("unhappy");
    }
    return 0;
}
/*
5 5 1
1 1 10
8
1 1 2
2 1 2
3 1 1
3 2 2

2 3 2
1 3 2
3 2 2
3 3 3


*/