poj 2391 (Floyd+最大流+二分)

题意：有n块草地，一些奶牛在草地上吃草，草地间有m条路，一些草地上有避雨点，每个避雨点能容纳的奶牛是有限的，给出通过每条路的时间，问最少需要多少时间能让所有奶牛进入一个避雨点。

两个避雨点间可以相互到达，所以必须要拆点，如果i-->j可以到达,加边i->j+n，流量无穷大，当然i->i+n也必须有边，，，

Folyd要用long long，，，，，

 

#include<stdio.h>
#include<string.h>
const int N=410;
const int inf=0x3fffffff;
int gap[N],dis[N],head[N],num,start,end,ans,n;
__int64 map[N][N];
struct edge
{
	int st,ed,flow,next;
}E[90000];
struct node
{
	int w1,w2;
}P[210];
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,flow=0,min_dis=ans-1;
	for(i=head[u];i!=-1;i=E[i].next)
	{
		v=E[i].ed;
		if(E[i].flow>0)
		{			
			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);
	//printf("maxflow=%d\n",maxflow);
	return maxflow;
}
void makemap(__int64 D)
{
	memset(head,-1,sizeof(head));
	num=0;
	int i,j;
	for(i=1;i<=n;i++)
	{
		addedge(i,i+n,inf);
		addedge(start,i,P[i].w1);
		for(j=i+1;j<=n;j++)
		{
			if(map[i][j]<=D)
			{
				addedge(i,j+n,inf);
				addedge(j,i+n,inf);
			}
		}
		addedge(i+n,end,P[i].w2);
	}
}
int main()
{
	int i,j,k,x,y,w,m,sum;
	while(scanf("%d%d",&n,&m)!=-1)
	{
		start=0;end=n*2+1;ans=end+1;sum=0;
		for(i=1;i<=n;i++)
		{
			for(j=i+1;j<=n;j++)
				map[i][j]=map[j][i]=1000000000000;
			map[i][i]=0;
		}
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&P[i].w1,&P[i].w2);
			sum+=P[i].w1;
		}
		for(i=1;i<=m;i++)
		{
			scanf("%d%d%d",&x,&y,&w);
			if(map[x][y]>w)
				map[x][y]=map[y][x]=w;
		}
		for(k=1;k<=n;k++)
			for(i=1;i<=n;i++)
				for(j=1;j<=n;j++)
					if(map[i][j]>map[i][k]+map[k][j])
						map[i][j]=map[i][k]+map[k][j];
		__int64 L,R,mid,flag;
		L=0;flag=-1;R=0;
		for(i=1;i<=n;i++)
			for(j=i+1;j<=n;j++)
				if(R<map[i][j]&&map[i][j]<1000000000000)
					R=map[i][j];
		while(L<=R)
		{
			mid=(L+R)/2;
			makemap(mid);
			//printf("mid=%d\n",mid);
			if(isap()==sum)
			{
				flag=mid;
				R=mid-1;
			}
			else L=mid+1;
		}
		printf("%I64d\n",flag);
	}
	return 0;
}