HDU 3549 Flow problem

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 8144    Accepted Submission(s): 3790

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)

 

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第一次做网络流的题目，求一个最大流

用 Dinic 算法 ，有向图 ， 反向弧的容量设为 0 。

每次用 bfs 构建层次图， 若果 vis[ t ] == 1 的话证明图上还是有增广路 。

然后用 递归dfs 寻找增广路 。

dfs 过程除了当前结点  x 以外 ， 还需要传入 一个表示“ 目前为止所有弧的最小残量 ” 的 a , 

当 x 为汇点 或者 a = 0的时候终止dfs 过程 ，否则多路增广 。

有一个重要的优化 就是保存每一个结点 x 正在考虑的 弧 cur[x] , 以避免重复计算 。( dfs 更新增广路之前把eh 复制给 cur 就ok 了)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;
const int INF = 1e9 ;
const int N = 20;
const int M = 2010;

bool vis[N];
int eh[N],et[M],nxt[M],ef[M],ec[M],etot;
int s , t , n , m;
int d[N],cur[N];
void init(){
    memset( eh, -1 , sizeof eh) ;
    etot =0 ;
}
void addedge( int u , int v , int c ,int f){
    et[etot] = v ; nxt[etot] = eh[u]; ef[etot] = f; ec[etot] = c ; eh[u] = etot++;
    et[etot] = u ; nxt[etot] = eh[v]; ef[etot] = f; ec[etot] = 0 ; eh[v] = etot++;
}

bool bfs ()
{
    memset( vis , 0 , sizeof vis);
    queue< int >que;
    que.push(s) ;
    d[s] = 0;
    vis[s] =1 ;
    while( !que.empty() ) {
        int u = que.front(); que.pop();
        for( int i = eh[ u ] ; ~i ; i = nxt[i] ){
            int v = et[i] , c = ec[i] , f = ef[i];
            if( !vis[v] && c > f){
                vis[v] = 1 ;
                d[v] = d[u] + 1;
                que.push(v);
            }
        }
    }
    return vis[t];
}

int dfs (int x ,int a)
{
    if ( x == t || a == 0 ){
        return a ;
    }
    int flow = 0 , F;
    for( int &i = cur[x] ; ~i ; i = nxt[i] ){
        int v = et[i] , c = ec[i] , &f = ef[i];
        if( d[x] + 1 == d[v] && ( F = dfs (v, min( a, c - f))) > 0) {
            f += F;
            ef[ i ^ 1 ] -= F;
            flow += F;
            a -= F;
            if( a == 0 )break;
        }
    }
    return flow;
}

int MF(){
    int flow = 0 ;
    while( bfs() ){
        memcpy( cur , eh , sizeof eh );
        flow += dfs(s , INF);
    }
    return flow;
}

void run()
{
    int u,v,c;

    scanf("%d%d",&n,&m);

    init();

    while(m--){
        scanf("%d%d%d",&u, &v, &c);
        addedge( u ,v , c , 0 );
    }
    s = 1 , t = n;
    printf("%d
",MF());
}

int main()
{
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    int _ , cas = 1 ;
    scanf("%d",&_);
    while(_--){
        printf("Case %d: ",cas++);
        run();
    }
    return 0;
}