HDU 3549 网络流初步

本文探讨了网络流这一经典问题,并通过一个具体实例讲解如何使用Ford-Fulkerson方法中的EK算法来解决最大流问题。文章包含完整的C++代码实现,帮助读者更好地理解算法原理及其应用。

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Flow Problem
Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 12912    Accepted Submission(s): 6154


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

昨天看了一个下午的网络流,大体上懂了 Ford - Fulkerson 方法的EK算法,拿这题贯通一下

#include <cstdio>
#include <cstring>
#include <queue>
#include <climits>
using namespace std;
#define MAX_V 15*2
#define MAX_E 1000*2

int Pre[MAX_E]; 
bool Visit[MAX_E];
int Map[MAX_E][MAX_E];      
bool BFS(int s,int t,int V,int E)
{
    int cur;
    queue<int> Q;
    memset(Pre,0,sizeof Pre);
    memset(Visit,false,sizeof Visit);
    Visit[s] = true;   Q.push(s);   
    while(!Q.empty())
    {
        cur = Q.front();Q.pop();
        if(cur == t) return true;  
        for(int i=1;i<=V;++i)
        {
            if(!Visit[i] && Map[cur][i]) 
            {
                Q.push(i);
                Pre[i] = cur;
                Visit[i] = true;
            }
        }
    }
    return false;
}
int Max_flow(int s,int t,int V,int E)  
{
    int flow = 0;
    while(true)
    {
        if(!BFS(s,t,V,E)) return flow;  
        int Min = INT_MAX;
        for(int i=t;i!=s;i=Pre[i]) 
            Min = min(Min,Map[Pre[i]][i]);
        for(int i=t;i!=s;i=Pre[i])
        {
            Map[Pre[i]][i] -= Min;  
            Map[i][Pre[i]] += Min;  
        }
        flow += Min;
    }
}
void Handle(int V,int E)
{
    memset(Map,0,sizeof Map);
    for(int i = 1,u,v,c;i <= E;i++) 
    {
        scanf("%d %d %d",&u,&v,&c);
        Map[u][v] += c;              
    }
}
int main(void)
{
  // freopen("F:\\test.txt","r",stdin);
    int T;scanf("%d",&T);
    for(int i=1;i<=T;i++)
    {
        int V,E;
        scanf("%d %d",&V,&E);   //V<=15,E<=1000,C<=1000
        Handle(V,E);
        printf("Case %d: %d\n",i,Max_flow(1,V,V,E));
    }
}
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