标签：acm   hdu   网络流   

Drainage Ditches

5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10

50

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 205;
int n, m;
int cap[maxn][maxn];
int flow[maxn];
int pre[maxn];

int bfs(int s, int t) {
    memset(pre, -1, sizeof(pre));
    pre[s] = 0;
    flow[s] = INF;
    
    queue<int> que;
    que.push(s);
    while(!que.empty()) {
        int cur = que.front();
        que.pop();
        if(cur == t) break;		//找到增广路径 
        for(int i = 1; i <= m; i++) {
            if(i != s && cap[cur][i] > 0 && pre[i] == -1) {
                pre[i] = cur;	//记录前驱 
                flow[i] = min(cap[cur][i], flow[cur]);
                que.push(i);
            }
        }
    }
    if(pre[t] == -1) return -1;
    else return flow[t];
}

int ek(int s, int t) {
    int inc = 0;
    int ans = 0;
    while((inc = bfs(s, t)) != -1) {
        int k = t;
        while(k != s) {
            cap[pre[k]][k] -= inc;
            cap[k][pre[k]] += inc;
            k = pre[k];
        }
        ans += inc;
    }
    return ans;
}

int main() {
    while(cin >> n >> m) {
        memset(cap, 0, sizeof(cap));
        for(int i = 0; i < n; i++) {
            int u, v, w;
            scanf("%d %d %d", &u, &v, &w);
            if(u == v) continue;
            cap[u][v] += w;
        }
        cout << ek(1, m) << endl;
    }
    return 0;
}

Flow Problem

2
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1

Case 1: 1
Case 2: 2

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 205;
int n, m;
int cap[maxn][maxn];
int flow[maxn];
int pre[maxn];

int bfs(int s, int t) {
	memset(pre, -1, sizeof(pre));
	pre[s] = 0;
	flow[s] = INF;
	
	queue<int> que;
	que.push(s);
	while(!que.empty()) {
		int cur = que.front();
		que.pop();
		if(cur == t) break; //找到增广路径 
		for(int i = 1; i <= m; i++) {
			if(i != s && cap[cur][i] > 0 && pre[i] == -1) {
				pre[i] = cur;	//记录前驱 
				flow[i] = min(cap[cur][i], flow[cur]);
				que.push(i);
			}
		}
	}
	if(pre[t] == -1) return -1;
	else return flow[t];
}

int ek(int s, int t) {
	int inc = 0;
	int ans = 0;
	while((inc = bfs(s, t)) != -1) {
		int k = t;
		while(k != s) {
			cap[pre[k]][k] -= inc;
			cap[k][pre[k]] += inc;
			k = pre[k];
		}
		ans += inc;
	}
	return ans;
}

int main() {
	int T;
	scanf("%d", &T);
	for(int cas = 1; cas <= T; cas++) {
		cin >> m >> n;
		memset(cap, 0, sizeof(cap));
		for(int i = 0; i < n; i++) {
			int u, v, w;
			scanf("%d %d %d", &u, &v, &w);
			if(u == v) continue;
			cap[u][v] += w;
		}
		cout << "Case " << cas << ": " << ek(1, m) << endl;
	}
	return 0;
}

HDU - 1532 - Drainage Ditches && 3549 - Flow Problem （网络流初步）

标签：acm   hdu   网络流   

原文地址：http://blog.csdn.net/u014355480/article/details/45103877