/*最大权闭包经典对于一个点连汇点，边权为花费值，连源点，边权为赚的的值两个点有约束关系a->b，连单项边b->a;赚的值的和减去最大流就是解*/#include <cstdio>#include <iostream>#include <memory.h>#include<queue>#include<set>#include<ctime>#include<algorithm>#include<cmath>#include<vector>#define LL __int64using namespace std;const int maxn=65009;//为dinic求最大流模版 const LL inf=(1LL)<<60;struct edge { int v, next; LL val; } net[ 5000010 ]; int n,m;int level[maxn], Qu[maxn], out[maxn],next[maxn]; class Dinic { public: int end; Dinic() { end = 0; memset( next, -1, sizeof(next) ); } inline void insert( int x, int y, LL c) { net[end].v = y, net[end].val = c, net[end].next = next[x], next[x] = end ++; net[end].v = x, net[end].val = 0, net[end].next = next[y], next[y] = end ++; } bool BFS( int S, int E ) { memset( level, -1, sizeof(level) ); int low = 0, high = 1; Qu[0] = S, level[S] = 0; for( ; low < high; ) { int x = Qu[low]; for( int i = next[x]; i != -1; i = net[i].next ) { if( net[i].val == 0 ) continue; int y = net[i].v; if( level[y] == -1 ) { level[y] = level[x] + 1; Qu[ high ++] = y; } } low ++; } return level[E] != -1; } LL MaxFlow( int S, int E ){ LL maxflow = 0; for( ; BFS(S, E) ; ) { memcpy( out, next, sizeof(out) ); int now = -1; for( ;; ) { if( now < 0 ) { int cur = out[S]; for(; cur != -1 ; cur = net[cur].next ) if( net[cur].val && out[net[cur].v] != -1 && level[net[cur].v] == 1 ) break; if( cur >= 0 ) Qu[ ++now ] = cur, out[S] = net[cur].next; else break; } int u = net[ Qu[now] ].v; if( u == E ) { LL flow = inf; int index = -1; for( int i = 0; i <= now; i ++ ) { if( flow > net[ Qu[i] ].val ) flow = net[ Qu[i] ].val, index = i; } maxflow += flow; for( int i = 0; i <= now; i ++ ) net[Qu[i]].val -= flow, net[Qu[i]^1].val += flow; for( int i = 0; i <= now; i ++ ) { if( net[ Qu[i] ].val == 0 ) { now = index - 1; break; } } } else{ int cur = out[u]; for(; cur != -1; cur = net[cur].next ) if (net[cur].val && out[net[cur].v] != -1 && level[u] + 1 == level[net[cur].v]) break; if( cur != -1 ) Qu[++ now] = cur, out[u] = net[cur].next; else out[u] = -1, now --; } } } return maxflow; } }; int main(){ int ca,c,v,r,num,a,b; scanf("%d",&ca); for(int kk=1;kk<=ca;kk++) { Dinic my; int en=60000; int TT=200; LL sum=0; scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",&num); for(int j=1;j<=num;j++) { scanf("%d%d%d",&c,&v,&r); sum+=v; my.insert(0,i*200+j,v); my.insert(i*200+j,en,c); while(r--) { scanf("%d%d",&a,&b); my.insert(i*200+j,a*200+b,inf); //my.insert(i*200+j,en,c); } } } printf("Case #%d: %I64d\n",kk,sum-my.MaxFlow(0,en)); }return 0;}
