POJ 2987 最大流最小割最大权闭合图

基本的最小割问题做得也差不多了....

这题整死我了.... 

在Amber的论文里有这个解法，大家去看看吧。

其他的不说了。

这题把数据类型全都改成long long才过，瞪大眼睛找了一晚上，没找出trick.... 悲催.... 

如果你也WA了，不妨试试这种方法... PS：我连bool都改long long了= =|||

#include<iostream> #include<cstdio> #define INF 2100000000000560ll #define MN 5555 #define MM 555555 #define CC(a) memset( a,0,sizeof(a) ) #define FF(i,N) for( int i=0;i<N;i++ ) template<class T>inline void checkmin( T&a,T b ){ if( a>b||a==-1 ) a=b; } using namespace std; typedef long long ll; ll n,m,s,t,alloc,tot; struct edge{ ll pos,next,c; }E[MM]; ll cur[MN],gap[MN],dis[MN],pre[MN],head[MN]; void addEdge( ll u,ll v,ll c,ll cc=0 ) { E[alloc].pos=v; E[alloc].c=c; E[alloc].next=head[u]; head[u]=alloc++; E[alloc].pos=u; E[alloc].c=cc; E[alloc].next=head[v]; head[v]=alloc++; } void setG() { ll u,v,cap; CC(E); memset( head,-1,sizeof(head) ); s=0,t=n+1;alloc=0;tot=0; for( int i=1;i<=n;i++ ){ scanf( "%lld",&cap ); if( cap>0 ){ addEdge( s,i,cap ); tot+=cap; } if( cap<0 ) addEdge( i,t,-cap ); } FF(i,m){ scanf( "%d%d",&u,&v ); addEdge( u,v,INF ); } } ll sap() { CC(dis),CC(gap); FF( i,t+1 ) cur[i]=head[i]; int u=pre[s]=s; ll maxflow=0,aug=-1; gap[0]=t+1; while( dis[s]<=t ){ loop: for( ll &i=cur[u];i!=-1;i=E[i].next ) { ll v=E[i].pos; if( E[i].c&&dis[u]==dis[v]+1 ) { pre[v]=u; checkmin( aug,E[i].c ); u=v; if( v==t ) { maxflow+=aug; //printf( "%d\n",maxflow ); for( u=pre[u];v!=s;v=u,u=pre[u] ) E[cur[u]].c-=aug,E[cur[u]^1].c+=aug; aug=-1; } goto loop; } } ll mind=t; for( ll i=head[u];i!=-1;i=E[i].next ) { ll v=E[i].pos; if( E[i].c&&mind>dis[v] ) cur[u]=i,mind=dis[v]; } if( --gap[dis[u]]==0 ) break; gap[dis[u]=mind+1]++; u=pre[u]; } return maxflow; } ll sum,vis[MN]; void dfs( ll cur ) { if( cur==t ) return ; sum++; vis[cur]=true; for( ll i=head[cur];i!=-1;i=E[i].next ) if( vis[E[i].pos]==false&&E[i].c ) dfs(E[i].pos); } int main() { while( scanf("%d%d",&n,&m)!=EOF ) { setG(); ll ans=tot-sap(); sum=0;CC(vis); dfs(s); printf( "%lld %lld\n",sum-1,ans ); } return 0; }