链接:click here
代码:
#include <ctype.h> //最大流 入门
#include <stdio.h>
#include <vector>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1101;
const int inf=0x3f3f3f3f;
struct edge {int to,cap,rev;};
vector<edge > G[maxn]; //图的邻接表表示
bool used[maxn]; //访问标记
void add_edge(int from,int to,int cap)
{
G[from].push_back((edge){to,cap,G[to].size()});
G[to].push_back((edge){from,0,G[from].size()-1});
}
int dfs(int v,int t,int f)
{
if(v==t) return f;
used[v]=true;
for(int i=0;i<G[v].size();i++)
{
edge &e=G[v][i];
if(!used[e.to]&&e.cap>0)
{
int d=dfs(e.to,t,min(f,e.cap));
if(d>0)
{
e.cap-=d;
G[e.to][e.rev].cap+=d;
return d;
}
}
}
return 0;
}
int max_flow(int s,int t)
{
int flow=0;
for(;;)
{
memset(used,0,sizeof(used));
int f=dfs(s,t,inf);
if(f==0) return flow;
flow+=f;
}
}
int main()
{
int n,m,too,capp,revv;
while(~scanf("%d%d",&n,&m))
{
memset(G,0,sizeof(G));
for(int i=0;i<n;i++)
{
scanf("%d%d%d",&too,&capp,&revv);
add_edge(too,capp,revv);
}
printf("%d\n",max_flow(1,m));
}
return 0;
}
#include <cstdio>
#include <vector>
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
const int N = 300;
const int MAX = 0x3f3f3f3f;
int map[N][N];
int flow[N][N];
int a[N],p[N];
int Ford_fulkerson(int s,int t)
{
queue<int> qq;
memset(flow,0,sizeof(flow));
int f=0,u,v;
while(1)
{
memset(a,0,sizeof(a));
a[s]=MAX;
qq.push(s);
while(!qq.empty())
{
u=qq.front();qq.pop();
for(v=1;v<=t;v++)
{
if(!a[v]&&map[u][v]>flow[u][v])//找到新结点v
{
p[v]=u;qq.push(v);//记录v的父亲,并加入FIFO队列
a[v]=a[u]<map[u][v]-flow[u][v]?a[u]:map[u][v]-flow[u][v];//a[v]为s-v路径上的最小流量
}
}
}
if(a[t]==0)
return f;
for(int i=t;i!=s;i=p[i])
{
flow[i][p[i]]-=a[t];
flow[p[i]][i]+=a[t];
}
f+=a[t];
}
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
memset(map,0,sizeof(map));
for(int i=0;i<n;i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
map[x][y]+=z;
}
printf("%d\n",Ford_fulkerson(1,m));
}
} NYOJ 323 && HDU 1532 && POJ 1273 Drainage Ditches (网络流之最大流入门)
原文地址:http://blog.csdn.net/u013050857/article/details/43372657