HDU-1532（网络最大流）

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 41 2 401 4 202 4 202 3 303 4 10

Sample Output

50

/* * 题意： 给你n条边和m个点，问你1-m的网络最大流是多少？ 题解： EK（Edmond—Karp）算法 */#include <queue>  #include<string.h>  #include<iostream>using namespace std;#define INF 0x3f3f3f3fint mp[205][205];         //记录残留网络的容量  int flow[205];  //标记从源点到当前节点实际还剩多少流量可用  int pre[205];             //标记在这条路径上当前节点的前驱,同时标记该节点是否在队列中  queue<int>Q;int n, m;int BFS(int Start, int End){while (!Q.empty()){Q.pop();}//初始化前置点for (int i = 1; i <= m; i++){pre[i] = -1;}pre[Start] = 0;flow[Start] = INF;     //初始的时候最大流设置为无限大Q.push(Start);while (!Q.empty()){int Index = Q.front();Q.pop();if (Index == End){break;}for (int i = 1; i <= n; i++){//如果当前剩余流量大于0，并且之前没有访问过该点if (i != Start && mp[Index][i] > 0 && pre[i] == -1){pre[i] = Index;   //当前点的前置是Index//更新源点到i点剩余的流量flow[i] = min(mp[Index][i], flow[Index]);Q.push(i);}}}// 如果不可以到达最后的点if (pre[End] == -1){return -1;}else{//返回起点到最后点剩余的流量return flow[End];}}int MaxFlow(int Start, int End)//返回 start -> End 的最大流{int ans = 0;int increaseflow = 0;//当前是否可以走到最后while ((increaseflow = BFS(Start, End)) != -1){int k = End;//下面代码用来修改地图流量while (k != Start){int Front_Index = pre[k];//Front_Index为k点的上一个点，利用前驱寻找路径mp[Front_Index][k] -= increaseflow;//改变正向边的容量  mp[k][Front_Index] += increaseflow;//改变反向边的容量  k = Front_Index;}ans += increaseflow;}return ans;}int main(){while (cin >> n >> m){int Start, End;int Val;memset(flow, 0, sizeof(flow));memset(mp, 0, sizeof(mp));for (int i = 0; i < n; i++){cin >> Start >> End >> Val;//如果起点终点相同则continueif (Start == End)continue;mp[Start][End] += Val;}cout << MaxFlow(1, m) << endl;}return 0;}