http://acm.hdu.edu.cn/showproblem.php?pid=4280
2 5 7 3 3 3 0 3 1 0 0 4 5 1 3 3 2 3 4 2 4 3 1 5 6 4 5 3 1 4 4 3 4 2 6 7 -1 -1 0 1 0 2 1 0 1 1 2 3 1 2 1 2 3 6 4 5 5 5 6 3 1 4 6 2 5 5 3 6 4
9 6
/**
hdu 4280 网络流+挂
题目大意:给出一些点从最西边的点经过网络到最东边的点,问最大流量(无向图)
解题思路:无向图网络流,不过无向图建边要正反向都要赋予流量值,如果有向图则正向流量值,反向为0;
*/
#pragma comment(linker, "/STACK:10240000000000,10240000000000")
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
const int oo=1e9;
const int mm=200051;
const int mn=100005;
int node,src,dest,edge;
int ver[mm],flow[mm],_next[mm];
int head[mn],work[mn],dis[mn],q[mn];
void prepare(int _node,int _src,int _dest)
{
node=_node,src=_src,dest=_dest;
for(int i=0; i<node; ++i)head[i]=-1;
edge=0;
}
void addedge(int u,int v,int c)
{
ver[edge]=v,flow[edge]=c,_next[edge]=head[u],head[u]=edge++;
ver[edge]=u,flow[edge]=c,_next[edge]=head[v],head[v]=edge++;
}
bool Dinic_bfs()
{
int i,u,v,l,r=0;
for(i=0; i<node; ++i)dis[i]=-1;
dis[q[r++]=src]=0;
for(l=0; l<r; ++l)
for(i=head[u=q[l]]; i>=0; i=_next[i])
if(flow[i]&&dis[v=ver[i]]<0)
{
dis[q[r++]=v]=dis[u]+1;
if(v==dest)return 1;
}
return 0;
}
int Dinic_dfs(int u,int exp)
{
if(u==dest)return exp;
for(int &i=work[u],v,tmp; i>=0; i=_next[i])
if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
{
flow[i]-=tmp;
flow[i^1]+=tmp;
return tmp;
}
return 0;
}
int Dinic_flow()
{
int i,ret=0,delta;
while(Dinic_bfs())
{
for(i=0; i<node; ++i)work[i]=head[i];
while(delta=Dinic_dfs(src,oo))ret+=delta;
}
return ret;
}
int n,m;
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
int a,b;
int maxx=-1,minn=0x3f3f3f3f;
for(int i=1;i<=n;i++)
{
int x,y;
scanf("%d%d",&x,&y);
if(maxx<x)
{
maxx=x;
b=i;
}
if(minn>x)
{
minn=x;
a=i;
}
}
prepare(n+1,a,b);
for(int i=0;i<m;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
}
int ans=Dinic_flow();
printf("%d\n",ans);
}
return 0;
}
原文:http://blog.csdn.net/lvshubao1314/article/details/44589661