POJ 2289 二分最大流

这中类型的题做过两道了，所以一见题就知道思路了。不过这题的输入比较恶心，还好能应付。

本来不想做这题的，但是看到数据也还大1500个点，想用链表试试，先敲了个二维数组，等待TLE的过程中1Y了....

何其蛋疼！1500ms 不算快...

#include<iostream>#include<cstdio>#include<cstdlib>#include<string>#include<map>#define CC(a) memset(a,0,sizeof(a))#define FF(i,a) for( int i=0;i<a;i++ )#define MN 1555#define ME 555555template<class T> void inline checkmin( T &a,T b ){ if(a==-1||a>b)a=b; }using namespace std;struct Edge{   int v,c,next;   }E[ME];int N,NE,M,s,t,CAN;int head[MN],gap[MN],pre[ME],dis[MN],cur[MN];int maze[MN][MN],cap[MN][MN];void setG(){  CC(cap);  map<string,int>map;  string str;int num;char c;  s=0;t=N+M+1;  FF( i,N ){   cin>>str;   map[str]=M+i+1;   num=-1;// people:[M+1,M+N] list:[1,M]   while( scanf("%c",&c) )   {  if( c==' '||c=='\n' ){ if( num==-1 ) continue; else cap[num+1][map[str]]++;if( c=='\n' ) break;num=-1;}else{ if( num==-1 ) num=c-'0'; else num=num*10+c-'0'; } } } for( int i=M+1;i<=M+N;i++ )   cap[i][t]=1;}void initG( int mid ){  for( int i=0;i<=t;i++ )  for( int j=0;j<=t;j++ )    maze[i][j]=cap[i][j];  for( int i=1;i<=M;i++ )    maze[s][i]=mid;}bool sap(){  CC(cur);CC(gap);CC(pre);CC(dis);  int u=pre[s]=s,maxflow=0,aug=-1;  gap[0]=t+1;  while( dis[s]<=t ){loop: for( int v=cur[u];v<=t;v++ ) if( maze[u][v]&&dis[u]==dis[v]+1 ) {   pre[v]=u;   cur[u]=v;   checkmin(aug,maze[u][v]);   u=v;   if( v==t )   {   maxflow+=aug;   for( u=pre[u];v!=s;v=u,u=pre[u] )   {    maze[u][v]-=aug;    maze[v][u]+=aug;  }  aug=-1;   }   goto loop;   }   int mind=t;   for( int v=0;v<=t;v++ )   if( maze[u][v]&&mind>dis[v] )   {   cur[u]=v;   mind=dis[v];   }   if( --gap[dis[u]]==0 )break;   gap[dis[u]=mind+1]++;   u=pre[u]; } if( maxflow==N ) return true; else return false;}int main(){ while( scanf("%d%d",&N,&M)!=EOF ) {    if( N==0&&M==0 )break;    setG();    int l=1,r=N,m;    while( (m=(l+r)/2)&&l<r )    {     initG(m);     if( sap() ) r=m;     else l=m+1;     }     printf( "%d\n",m );  } return 0;}