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;
}