/* 题目: 求最小费用最大流的裸模板 分析: 每次找到一条最短路,然后记录下每个经过的节点的前驱结点,然后若能够 到达终点的话,从终点开始往回找到最小流,然后再从终点往回依次更改当 前每条边的流量以及反向弧,直到不能再找到一条最短路为止,找最短路的 算法是spfa */ #include <iostream> #include <cstdio> #include <cstring> using namespace std; const int maxn = 10005; const int maxm = 1000005; const int inf = 1e8; #define debug puts("here"); int n,m,s,t,cas,ans,cl,aug; int dis[maxn],pre[maxn],po[maxn]; bool use[maxn]; int q[maxm],head,tail; struct node { int y,next,f,c; }edge[maxm]; void add(int x,int y,int f,int c) { edge[++cl].y = y; edge[cl].f = f; edge[cl].c = c; edge[cl].next = po[x]; po[x] = cl; edge[++cl].y = x; edge[cl].f = 0; edge[cl].c = -c; edge[cl].next = po[y]; po[y] = cl; } int spfa() { memset(use,false,sizeof(use)); for(int i=0;i<=n;i++) dis[i] = inf; head = tail = 0; q[tail++] = s; dis[s] = 0; pre[s] = 0; int x,y; while(head<tail) { x = q[head++]; use[x] = false; for(int i=po[x];i;i=edge[i].next) { y = edge[i].y; if(edge[i].f>0&&edge[i].c+dis[x]<dis[y]) { dis[y] = edge[i].c+dis[x]; pre[y] = i; if(!use[y]) { q[tail++] = y; use[y] = true; } } } } if(dis[t]==inf) return false; aug = inf; for(int i=pre[t];i;i=pre[edge[i^1].y]) aug = min(aug,edge[i].f); for(int i=pre[t];i;i=pre[edge[i^1].y]) { edge[i].f -= aug; edge[i^1].f += aug; } ans += dis[t]*aug; return true; } int main() { freopen("sum.in","r",stdin); int ncase,x,y,f,c; char str[10]; cin>>ncase; while(ncase--) { scanf("%s%d",str,&x); scanf("%d%d%d%d",&n,&m,&s,&t); memset(po,0,sizeof(po)); cl = 1; for(int i=0;i<m;i++) { scanf("%d%d%d%d",&x,&y,&f,&c); add(x,y,f,c); } ans = 0; while(spfa()); cout<<ans<<endl; } return 0; }