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，
多加一个点一定是y增加比较快_(:зゝ∠)_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
*/
HDU 4940 Destroy Transportation system 规律题,布布扣,bubuko.com
HDU 4940 Destroy Transportation system 规律题
原文：http://blog.csdn.net/qq574857122/article/details/38518189