Drainage Ditches(网络流_EK模板)
来源:互联网 发布:西安丝路软件 编辑:程序博客网 时间:2024/06/05 09:06
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.
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
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;}
0 0
- Drainage Ditches(网络流_EK模板)
- hdu 1532 Drainage Ditches(网络流dinic模板)
- hdu1532——Drainage Ditches(网络流模板)
- usaco Drainage Ditches(网络流dinic模板)
- 网络流模板-(EdmondsKarp)-HDU-1532-Drainage Ditches
- HDU1532 Drainage Ditches(网络流模板)
- 网络流模板 POJ 1273 Drainage Ditches
- POJ_P1273 Drainage Ditches(模板题+网络流)
- HDU 1532 Drainage Ditches 网络流模板
- HDU1532 Drainage Ditches 网络流模板
- poj1273 Drainage Ditches (最大流模板)
- POJ 1273 - Drainage Ditches (网络流)
- Drainage Ditches(HDU1532,网络流)
- POJ 1273 Drainage Ditches (网络流)
- POJ 1273Drainage Ditches --网络流最大流模板题
- 【网络流】最大流:Drainage Ditches【EK模板】
- POJ1273 Drainage Ditches(网络流最大流模板)
- HDU 1532 Drainage Ditches 网络流 模板题
- HDU 2094 STL
- 10.获取客户端指定路径下的文件目录,并传输至服务器端--客户端
- AM使用指南之七:Configuration参数说明(1)
- 快速排序的简单实现
- Libgdx的使用(15)——使用Gradle构建速度慢的问题
- Drainage Ditches(网络流_EK模板)
- AM使用指南之七:Configuration参数说明(2)
- OpenERP 之 Relational Types(关联类型)
- AM使用指南之七:Configuration参数说明(3)
- 串的模式匹配算法(求子串位置的定位函数Index(S,T,pos))
- 10.获取客户端指定路径下的文件目录,并传输至服务器端--服务器端
- 开博客了
- cocos2dx 运动+旋转动画 CCSequence CCAnimation CCAnimate CCMoveTo CCCallFuncN
- 黑马程序员,黑马论坛------Scanner的小问题(大家来找茬)