【HDOJ】3549 Flow Problem

网络流基础题目，Edmonds_Karp可解。

/* 3549 */
#include <iostream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
using namespace std;

#define rep(i, a, n)     for (int i=a;i<n;++i)
#define per(i, a, n)     for (int i=n-1;i>=a;--i)
#define pb                 push_back
#define mp                 make_pair
#define all(x)             (x).begin(),(x).end()
#define SZ(x)             ((int)(x).size())
#define lson            l, mid, rt<<1
#define rson            mid+1, r, rt<<1|1

const int maxn = 20;
int F[maxn][maxn];
int P[maxn], a[maxn];
int n, m, ans;
int s, t;

bool bfs() {
    int u, v;
    queue<int> Q;
    
    memset(a, 0, sizeof(a));
    a[s] = INT_MAX;
    Q.push(s);
    P[s] = s;
    
    while (!Q.empty()) {
        u = Q.front();
        Q.pop();
        for (v=1; v<=n; ++v) {
            if (!a[v] && F[u][v]) {
                P[v] = u;
                Q.push(v);
                a[v] = min(a[u], F[u][v]);
            }
        }
    }
    
    return a[t] == 0;
}

void Edmonds_Karp() {
    int u, v;
    
    s = 1;
    t = n;
    ans = 0;
    
    while (true) {
        if (bfs())
            break;
        for (u=t; u!=s; u=P[u]) {
            F[u][P[u]] += a[t];
            F[P[u]][u] -= a[t];
        }
        ans += a[t];
    }
}

int main() {
    int i, j, k;
    int t, tt;
    
    #ifndef ONLINE_JUDGE
        freopen("data.in", "r", stdin);
        freopen("data.out", "w", stdout);
    #endif
    
    scanf("%d", &tt);
    for (t=1; t<=tt; ++t) {
        scanf("%d %d", &n, &m);
        memset(F, 0, sizeof(F));
        while (m--) {
            scanf("%d %d %d", &i, &j, &k);
            F[i][j] += k;
        }
        Edmonds_Karp();
        printf("Case %d: %d
", t, ans);
    }
    
    #ifndef ONLINE_JUDGE
        printf("time = %d.
", (int)clock());
    #endif
    
    return 0;
}