Drainage Ditches（网络流_EK模板）

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

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

Sample Output

50

解题思路

直接的模板题，下面的代码还是比较好理解的，寻找增广路，能找到就找出这条增广路的最小流（瓶颈），更新g[]数组，继续找增广路，找不到增广路或源点汇点重合时

结束。

AC代码

#include <cstdio>
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
const int maxn = 205;
const int INF = (1 << 30) - 1;
int g[maxn][maxn];
int flow[maxn], pre[maxn];
bool vis[maxn];
int n, m;

int bfs(int s, int e)
{
    memset(pre, -1, sizeof(pre));
    memset(vis, false, sizeof(vis));
    queue<int> q;
    vis[s] = true;
    for(int i = 1; i <= n; i++)
        flow[i] = INF;
    q.push(s);
    while(!q.empty())
    {
        int now = q.front();
        q.pop();
        if(now == n)
            break;
        for(int i = 1; i <= n; i++)
        {
            if(!vis[i] && g[now][i] > 0)
            {
                vis[i] = true;
                flow[i] = min(flow[now], g[now][i]);
                pre[i] = now;
                q.push(i);
            }
        }
    }
    if(!vis[e] || e == 1)
        return -1;
    else
        return flow[e];
}

int edmonds_karp(int s, int e)
{
    int temp, d, res, maxflow;
    maxflow = 0;
    while( (d = bfs(s, e)) != -1 )
    {
        maxflow += d;
        temp = n;
        while(temp != 1)
        {
            res = pre[temp];
            g[res][temp] -= d;
            g[temp][res] += d;
            temp =res;
        }
    }
    return maxflow;
}

int main()
{
    int start, end, capacity;
    while(scanf("%d%d", &m, &n) != EOF)
    {
        memset(g, 0, sizeof(g));
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d%d", &start, &end, &capacity);
            g[start][end] += capacity;
        }
        printf("%d\n", edmonds_karp(1, n));
    }
    return 0;
}