poj 2112 (最大流+二分)

题意：有k台挤奶机，c头奶牛，给出这k+c个实体间的距离，求出每头奶牛都到一台挤奶机去，怎么分配使奶牛走的最大距离最小。

用二分枚举最大距离，，，，

#include
#include
#define N 500
#define inf 0x3fffffff
int map[N][N],dis[N],gap[N],head[N],num,n,m,D,start,end,ans;
struct edge
{
	int st,ed,flow,next;
}E[N*40];
void addedge(int x,int y,int w)
{
	E[num].st=x;E[num].ed=y;E[num].flow=w;E[num].next=head[x];head[x]=num++;
	E[num].st=y;E[num].ed=x;E[num].flow=0;E[num].next=head[y];head[y]=num++;
}
int dfs(int u,int minflow)
{
	if(u==end)return minflow;
	int i,v,f,min_dis=ans-1,flow=0;
	for(i=head[u];i!=-1;i=E[i].next)
	{
		if(E[i].flow>0)
		{
			v=E[i].ed;
			if(dis[v]+1==dis[u])
			{
				f=dfs(v,E[i].flow>minflow-flow minflow-flow:E[i].flow);
				E[i].flow-=f;
				E[i^1].flow+=f;
				flow+=f;
				if(flow==minflow)break;
				if(dis[start]>=ans)return flow;
			}
			min_dis=min_dis>dis[v] dis[v]:min_dis;
		}
	}
	if(flow==0)
	{
		if(--gap[dis[u]]==0)
			dis[start]=ans;
		dis[u]=min_dis+1;
		gap[dis[u]]++;
	}
	return flow;
}
int isap()
{
	int maxflow=0;
	memset(dis,0,sizeof(dis));
	memset(gap,0,sizeof(gap));
	gap[0]=ans;
	while(dis[start]map[i][k]+map[k][j])
							map[i][j]=map[i][k]+map[k][j];
						if(right