hdu 3549 Flow Problem(最大流 EK,sap)

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

题意：首先给定有多少组测试数据，然后给定一个顶点数，边数，然后给定一个单向的流量，求从第一个点到最后一个点的最大流量。注意该流网络是有向图；

最大流基础题，可以ek或者sap算法；

代码：

ek算法

#include
#include
#include
#include
using namespace std;
int n,m;
const int N=20;
int map[N][N];
int pre[N],vis[N];
int bfs(int s,int t)
{
    memset(pre,0,sizeof(pre));
    memset(vis,0,sizeof(vis));
    queueque;
    que.push(s);
    while(que.size())
    {
        int q=que.front();
        que.pop();
        if(q==t)
            return 1;
        for(int i=1;i<=t;i++)
        {
            if(map[q][i]&&!vis[i])
            {
                pre[i]=q;
                vis[i]=1;
                que.push(i);
            }
        }
    }
    return 0;
}
int ek(int s,int t)
{
    int min=0;
    while(bfs(s,t))
    {
        int minn=0xfffffff;
        for(int i=t;i!=s;i=pre[i])
        {
            if(minn>map[pre[i]][i])
                minn=map[pre[i]][i];
        }
        min+=minn;
        for(int i=t;i!=s;i=pre[i])
        {
            map[pre[i]][i]-=minn;
            map[i][pre[i]]+=minn;
        }
    }
    return min;
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int t=1;t<=T;t++)
    {
        scanf("%d%d",&n,&m);
        memset(map,0,sizeof(map));
        for(int i=1;i<=m;i++)
        {
            int a,b,w;
            scanf("%d%d%d",&a,&b,&w);
            map[a][b]+=w;
        }
        printf("Case %d: ",t);
        printf("%d\n",ek(1,n));
    }
}

sap

#include
#include
#include
using namespace std;
const int N=20;
int leve[N];
int n,m;
int map[N][N],gap[N],pre[N];
int sap(int s,int t)
{
    memset(pre,0,sizeof(pre));
    memset(gap,0,sizeof(gap));
    memset(leve,0,sizeof(leve));
    int u;
    int min=0;
    u=pre[s]=s;
    gap[0]=t;
    while(leve[s]map[pre[i]][i])
                        minn=map[pre[i]][i];
                min+=minn;
                for(int i=t;i!=s;i=pre[i])
                {
                    map[pre[i]][i]-=minn;
                    map[i][pre[i]]+=minn;
                }
                u=s;
            }
        }
        else
        {
            int minleve=t;
            for(int i=1;i<=n;i++)
                 if(minleve>leve[i]&&map[u][i])
                     minleve=leve[i];
            gap[leve[u]]--;
            if(gap[leve[u]]==0)
                break;
            leve[u]=minleve+1;
            gap[leve[u]]++;
            u=pre[u];
        }
    }
    return min;
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int t=1;t<=T;t++)
    {
        scanf("%d%d",&n,&m);
        memset(map,0,sizeof(map));
        for(int i=1;i<=m;i++)
        {
            int a,b,w;
            scanf("%d%d%d",&a,&b,&w);
            map[a][b]+=w;
        }
        printf("Case %d: ",t);
        printf("%d\n",sap(1,n));
    }
}