【bzoj1305】[CQOI2009]dance跳舞 最大流+二分答案

二分答案
每个点拆成两个点x,x'
每个男生x向x'连一条容量为k的边
每个女生y'向y连一条容量为k的边
源点S向每个男生连一条容量为ans的边
每个女生向汇点T连一条容量为ans的边
对于男生x和女生y，
如果互相喜欢，则x向y连一条容量为1的边
如果不互相喜欢，则x’向y'连一条容量为1的边
若最大流为n*ans则可行，否则不可行
点数4n+2

边数(4n+n^2)*2

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#define inf 1000000000
#define maxn 310
#define maxm 12010

using namespace std;

int head[maxn],to[maxm],next[maxm],c[maxm],q[maxn],d[maxn];
int n,m,num,s,t,k;
char s1[maxn];
int a[maxn][maxn];

void addedge(int x,int y,int z)
{
	num++;to[num]=y;c[num]=z;next[num]=head[x];head[x]=num;
	num++;to[num]=x;c[num]=0;next[num]=head[y];head[y]=num;
}

bool bfs()
{
	memset(d,-1,sizeof(d));
	int l=0,r=1;
	q[1]=s;d[s]=0;
	while (l<r)
	{
		int x=q[++l];
		for (int p=head[x];p;p=next[p])
		  if (c[p] && d[to[p]]==-1)
		  {
		  	d[to[p]]=d[x]+1;
		  	q[++r]=to[p];
		  }
	}
	if (d[t]==-1) return 0;
	else return 1;
}

int find(int x,int low)
{
	if (x==t || low==0) return low;
	int totflow=0;
	for (int p=head[x];p;p=next[p])
	  if (c[p] && d[to[p]]==d[x]+1)
	  {
	  	int a=find(to[p],min(low,c[p]));
	  	c[p]-=a;c[p^1]+=a;
	  	totflow+=a;low-=a;
	  	if (low==0) return totflow;
	  }
	if (low) d[x]=-1;
	return totflow;
}

bool Dinic(int x)
{
	num=1;
	memset(head,0,sizeof(head));
	memset(c,0,sizeof(c));
	s=0;t=4*n+1;
	for (int i=1;i<=n;i++) addedge(s,i,x),addedge(i,n+i,k),addedge(2*n+i,3*n+i,k),addedge(3*n+i,t,x);
	for (int i=1;i<=n;i++)
	  for (int j=1;j<=n;j++)
	    if (a[i][j]) addedge(i,3*n+j,1);
	    else addedge(n+i,2*n+j,1);
	int ans=0;
	while (bfs()) ans+=find(s,inf);
	if (ans==n*x) return 1;
	else return 0;
}

int main()
{
	scanf("%d%d",&n,&k);
	for (int i=1;i<=n;i++)
	{
		scanf("%s",s1+1);
		for (int j=1;j<=n;j++)
		  if (s1[j]=='Y') a[i][j]=1; else a[i][j]=0;
	}
	int l=1,r=n,ans=0;
	while (l<=r)
	{
		int mid=(l+r)/2;
		if (Dinic(mid)) l=mid+1,ans=mid;
		else r=mid-1;
	}
	printf("%d\n",ans);
	return 0;
}