HDU 3987 Harry Potter and the Forbidden Forest 最小割

/*很经典的最大流最小割的题目题意：求最小割，但因为最小割是不唯一的，题目要求得到最小割的条件下使得割边最少搜到usaco类似的一个题目才出的，构造很巧妙建边的时候每条边权 w=w*(E+1)+1;这样得到最大流maxflow/(E+1) 就是答案了道理很简单，如果原先两类割边都是最小割，那么求出的最大流相等但边权变换后只有边数小的才是最小割了，至于为什么乘的是（E+1）是为了保证边数叠加后依然是余数，不至于影响求最小割的结果*/#include <cstdio>#include <iostream>#include <memory.h>#include<queue>using namespace std;const int maxn=1009;const __int64 inf=(1LL)<<60;typedef  __int64   LL;//**************************************************  //为dinic求最大流模版  struct edge  {       int v, next;       LL val;   } net[ 500010 ];   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,t,u,v;    LL w;    scanf("%d",&ca);    for(int k=1;k<=ca;k++)    {        Dinic my;        scanf("%d%d",&n,&m);        while(m--)        {            scanf("%d%d%I64d%d",&u,&v,&w,&t);            my.insert(u,v,w*100001+1);            if(t==1)            {                my.insert(v,u,w*100001+1);            }        }        LL ans=my.MaxFlow(0,n-1);        //cout<<ans<<endl;        printf("Case %d: %I64d\n",k,(ans%100001));    }    return 0;}