poj 2455(二分+最大流)

题意：从1到n，要走不少于K条路，每条路不能重复走，点可以重复走，求所走的路中最长的最小，，

找出距离的最大值，然后二分，建图时符合的每条边链接的两点间建双向边，流量都为1，求最大流

#include<stdio.h>
#include<string.h>
const int N=210;
const int inf=0x3fffffff;
int dis[N],gap[N],head[N],num,start,end,ans,n,m;
struct edge
{
	int st,ed,flow,next;
}E[200000];
struct node
{
	int x,y,w;
}P[40000];
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=w;E[num].next=head[y];head[y]=num++;
}
void makemap(int D)
{
	memset(head,-1,sizeof(head));
	num=0;
	int i;
	for(i=0;i<m;i++)
	{
		if(P[i].w<=D)
			addedge(P[i].x,P[i].y,1);
	}
}
int dfs(int u,int minflow)
{
	if(u==end)return minflow;
	int i,v,f,flow=0,min_dis=ans-1;
	for(i=head[u];i!=-1;i=E[i].next)
	{
		if(E[i].flow)
		{
			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]<ans)
	   maxflow+=dfs(start,inf);
	return maxflow;
}
int main()
{
	int i,k,L,R,mid;
	while(scanf("%d%d%d",&n,&m,&k)!=-1)
	{
		R=0;start=1;end=n;ans=n;
		for(i=0;i<m;i++)
		{
			scanf("%d%d%d",&P[i].x,&P[i].y,&P[i].w);
			if(R<P[i].w)R=P[i].w;
		}
		L=0;
		while(L<R)
		{
           mid=(L+R)/2;
		   makemap(mid);
		   if(isap()>=k)
			   R=mid;
		   else L=mid+1;
		}
		printf("%d\n",R);
	}
	return 0;
}