3549 Flow Problem(入门) [最大流]

最新推荐文章于 2022-11-15 09:21:54 发布

最新推荐文章于 2022-11-15 09:21:54 发布 · 367 阅读

题解同时被 2 个专栏收录

23 篇文章

订阅专栏

网络流

7 篇文章

订阅专栏

本文介绍了一种解决网络流问题的方法，并提供了一个具体的C++实现示例。该算法通过不断寻找增广路径来逐步增加从源点到汇点的流量，直至无法找到新的增广路径为止。

Flow Problem

Time Limit: 5000 MS Memory Limit: 32768 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

[Submit] [Status] [Discuss]

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

同样一道网络流模板题，注意格式就行。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int MAXN = 30;
const int INF = 2147483647;
struct edge{
   int cap,to,rev;
   edge(int a = 0,int b = 0,int c = 0):to(a),cap(b),rev(c){}
};
vector<edge> G[MAXN];
int level[MAXN];
int iter[MAXN];

void Add(int f,int t,int c)
{
    G[f].push_back(edge{t,c,G[t].size()} );
    G[t].push_back(edge{f,0,G[f].size()-1} );
}

void bfs(int s)
{
    memset(level,-1,sizeof(level));
    level[s] = 0;
    queue<int > que;
    que.push(s);
    while(!que.empty()){
        int v = que.front();
        que.pop();
        for(int i = 0;i < G[v].size();i++){
            edge &e = G[v][i];
            if(e.cap > 0 && level[e.to] == -1){
                level[e.to] = level[v]+1;
                que.push(e.to);
            }
        }
    }
}

int dfs(int v,int t,int f)
{
    if(v == t) return f;
    for(int &i = iter[v];i < G[v].size();i++){
        edge &e = G[v][i];
        if(e.cap > 0 && level[v] < level[e.to]){
            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 maxd(int s,int t)
{
    int tempf = 0;
    while(true){
        bfs(s);
        if(level[t] < 0) return tempf;
        memset(iter,0,sizeof(iter));
        int f;
        while((f = dfs(s,t,INF)) > 0 ) tempf += f;
    }
}

void init()
{
    for(int i = 0;i < MAXN;i++){
        G[i].clear();
    }
}

int main()
{
    int t;
    int Case = 1;
    while(scanf("%d",&t) != EOF){
        while(t--){
            int dsum,bsum;
            init();
            scanf("%d%d",&dsum,&bsum);
            for(int i = 0;i < bsum;i++){
                int a,b,c;
                scanf("%d%d%d",&a,&b,&c);
                Add(a,b,c);
            }
            printf("Case %d: %d\n",Case++,maxd(1,dsum));
        }
    }
    return 0;
}