hdoj 1532 Drainage Ditches （最大流）

最新推荐文章于 2019-03-02 18:29:34 发布

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#最大流

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解决农场中雨水通过一系列排水沟流向溪流的问题，利用最大流算法确定水从池塘流向溪流的最大速率。

Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 53 Accepted Submission(s): 44

Problem Description

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.

Input

The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.

Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

Sample Input

Sample Output

Source

USACO 93

最大流的入门题，看的算法导论跟白书的讲解。。。但还是不理解源点到汇点用最短路优化是什么道理。。

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
#include <cmath>
#include <vector>
#include <utility>
#include <queue>
using namespace std;
const int INF = 0x3f3f3f3f;
int n, m;
struct Dinic {
    static const int maxn = 205;
    int g[maxn][maxn];
    int dis[maxn];//据原点的距离，分层图
    
    void init() {
        memset(g, 0, sizeof(g));
    }
    
    bool bfs() {
        memset(dis, -1, sizeof(dis));
        queue<int> que;
        dis[1] = 0;
        que.push(1);
        while (!que.empty()) {
            int u = que.front();
            que.pop();
            
            for (int i = 1; i <= n; ++i) {
                if (dis[i] < 0 && g[u][i] > 0) {
                    dis[i] = dis[u] + 1;
                    que.push(i);
                }
            }
        }
        
        if (dis[n] > 0)
            return true;
        else
            return false;
    }
    
    int find(int u, int low) {
        int res = 0;
        if (u == n) // x为汇点
            return low;
        
        for (int v = 1; v <= n; ++v) {
            int w = g[u][v];
            if (w > 0 && dis[u] + 1 == dis[v] && (res = find(v, min(w, low)))) {
                g[u][v] -= res;
                g[v][u] += res;
                return res;
            }
        }
        return 0;
    }
    
    int ans() {
        int res = 0;
        while (bfs()) {
            int temp;
            while ((temp = find(1, INF))) {
                res += temp;
            }
        }
        return res;
    }
}foo;
int main() {
        //freopen("in.txt", "r", stdin);
    while (cin >> m >> n) {
        foo.init();
        for (int i = 0; i < m; ++i) {
            int a, b, c;
            cin >> a >> b >> c;
            foo.g[a][b] += c;     //防止重边的存在
        }
        cout << foo.ans() << endl;
    }
}