hdoj3549 Flow Problem（裸的网络流）

本文介绍了一种经典的网络流问题，并提供了完整的解决方案。通过构建加权有向图，使用深度优先搜索寻找增广路径来求解最大流量问题。文章包含详细的算法步骤及C++实现代码。

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Flow Problem

Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 7647 Accepted Submission(s): 3560

Problem Description

Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.

Input

The first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)

Output

For each test cases, you should output the maximum flow from source 1 to sink N.

Sample Input

Sample Output

  
   Case 1: 1
Case 2: 2

Author

HyperHexagon

Source

HyperHexagon's Summer Gift (Original tasks)

第一道网络流题目

就是套模板

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include<stack>
using namespace std;

typedef long long LL;
const int N=22;
const int INF=100000000;
struct edge
{
int to,cap,res;
};
vector<edge> g[N];
int used[N];
void add_edge(int from,int to,int cap)
{
edge tmp;
tmp.to=to;
tmp.cap=cap;
tmp.res=g[to].size();
g[from].push_back(tmp);
tmp.to=from;
tmp.cap=0;
tmp.res=g[from].size()-1;
g[to].push_back(tmp);
}
int dfs(int s,int t,int f) //　找一条增广路
{
if(s==t) return f;
used[s]=1;
for(int i=0;i<g[s].size();i++)
{
edge &e=g[s][i];
int d;
if(!used[e.to]&&e.cap>0) // e.cap>0 要满足，不然搜下去会为负值
{
d= dfs(e.to,t,min(f,e.cap));
if(d>0)
{
e.cap-=d;
g[e.to][e.res].cap+=d;
return d;
}
}
}
return 0;
}
void init()
{
for(int i=0;i<=N;i++)
g[i].clear();
}
LL max_flow(int s,int t)
{
LL ans=0;
for(;;)
{
memset(used,0,sizeof(used));
LL d=dfs(s,t,INF);
if(d==0)
return ans;
ans+=d;
}
}
int main()
{
int ca,v_sum,e_sum;
scanf("%d",&ca);
for(int kase=1;kase<=ca;kase++)
{
scanf("%d%d",&v_sum,&e_sum);
init();
for(int i=0;i<e_sum;i++)
{
int aa,bb,cc;
scanf("%d%d%d",&aa,&bb,&cc);
add_edge(aa,bb,cc);
}
printf("Case %d: ",kase);
printf("%lld\n",max_flow(1,v_sum));
}
return 0;
}