[ACM] hdu 3549 Flow Problem （最大流模板题）

Flow Problem

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

2 3 2 1 2 1 2 3 1 3 3 1 2 1 2 3 1 1 3 1

 

Sample Output

Case 1: 1 Case 2: 2

 

Author

HyperHexagon

 

Source

HyperHexagon's Summer Gift (Original tasks)

解题思路：

本题为赤裸裸的最大流模板题。Ford-Fulkerson方法

其中为什么要用到残留网络，不好理解。

下面讲解有所启发，转载于：http://www.cnblogs.com/acSzz/archive/2012/09/13/2683820.html

必须使用反向弧的流网络

在这幅图中我们首先要增广1->2->4->6,这时可以获得一个容量为 2的流,但是如果不建立4->2反向弧的话,则无法进一步增广,最终答案为2,显然是不对的,然而如果建立了反向弧4->2,则第二次能进行 1->3->4->2->5->6的增广,最大流为3.

 

Comzyh对反向弧的理解可以说是"偷梁换柱",请仔细阅读: 在上面的例子中,我们可以看出,最终结果是1->2->5->6和1->2->4->6和 1->3->4->6.当增广完1->2->4->6(代号A)后,在增广 1->3->4->2->5->6(代号B),相当于将经过节点2的A流从中截流1(总共是2)走2->5>6,而不走2->4>6了,同时B流也从节点4截流出1(总共是1)走4->6而不是4->2->5->6,相当于AB流做加法.

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
using namespace std;
const int maxn=20;
const int inf=0x3f3f3f3f;
int pre[maxn];   //保存前驱节点
bool vis[maxn];
int mp[maxn][maxn];//临接矩阵保存残留网络

int s,e;//s为源点，e为汇点
int n,m;//输入n个顶点，m条边

bool bfs()
{
    queueq;
    memset(pre,0,sizeof(pre));
    memset(vis,0,sizeof(vis));
    vis[s]=1;
    q.push(s);
    while(!q.empty())
    {
        int first=q.front();
        q.pop();
        if(first==e)
            return true;//找到一条增广路
        for(int i=1;i<=n;i++)
        {
            if(!vis[i]&&mp[first][i])
            {
                q.push(i);
                pre[i]=first;
                vis[i]=1;
            }
        }
    }
    return false;
}

int max_flow()
{
    int ans=0;
    while(1)
    {
        if(!bfs())//找不到增广路
            return ans;
        int Min=inf;
        for(int i=e;i!=s;i=pre[i])//回溯找最小流量
            Min=min(Min,mp[pre[i]][i]);
        for(int i=e;i!=s;i=pre[i])
        {
            mp[pre[i]][i]-=Min;
            mp[i][pre[i]]+=Min;
        }
        ans+=Min;
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    int u,v,c;
    for(int cas=1;cas<=t;cas++)
    {
        scanf("%d%d",&n,&m);
        s=1,e=n;
        memset(mp,0,sizeof(mp));
        while(m--)
        {
            scanf("%d%d%d",&u,&v,&c);
            mp[u][v]+=c;
        }
        printf("Case %d: %d\n",cas,max_flow());
    }
    return 0;
}