zoukankan      html  css  js  c++  java
  • poj_1273Drainage Ditches

    Drainage Ditches
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 47664   Accepted: 17948

    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

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

    Sample Output

    50

    网络流第一题,参考刘汝佳的白书

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<map>
    #include<cmath>
    #include<vector>
    #include<algorithm>
    #include<set>
    #include<string>
    #include<queue>
    #include <stack>
    using namespace std;
    #pragma warning(disable : 4996)
    const int MAXN = 250;
    const int INF = 0x7FFFFFFF;
    int flow[MAXN][MAXN], cap[MAXN][MAXN];
    int maxflow, n, m;
    
    void EKarp(int s, int e)
    {
    	queue<int>Q;
    	int u, v, a[MAXN], father[MAXN];
    	maxflow = 0;
    	memset(flow, 0, sizeof(flow));
    	while (true)
    	{
    		memset(a, 0, sizeof(a));
    		a[s] = INF;
    		Q.push(s);
    		while (!Q.empty())//BFS找增广路
    		{
    			int u = Q.front();
    			Q.pop();
    			for (v = 1; v <= n; v++)
    			{
    				if(a[v] == 0 && cap[u][v] > flow[u][v])//找到新节点v
    				{
    					father[v] = u;//记录v的父亲,并加入FIFO队列
    					Q.push(v);
    					a[v] = min(a[u], cap[u][v] - flow[u][v]);//s-v路径上的最小残量
    				}
    			}
    		}
    		if(a[e] == 0)//找不到,则当前流已经是最大流
    		{
    			break;
    		}
    		for (u = e; u != s; u = father[u])//从汇点往回走
    		{
    			flow[father[u]][u] += a[e];//更新正向流量
    			flow[u][father[u]] -= a[e];//更新反向流量
    		}
    		maxflow += a[e];//更新从s流出的总流量
    	}
    }
    
    int main()
    {
    	freopen("in.txt", "r", stdin);
    	int x, y, z;
    	while(scanf("%d %d", &m, &n) != EOF)
    	{
    		memset(cap, 0, sizeof(cap));
    		while (m--)
    		{
    			scanf("%d %d %d", &x, &y, &z);
    			cap[x][y] += z;
    		}
    		EKarp(1, n);
    		printf("%d\n", maxflow);
    	}
    	return 0;
    }


  • 相关阅读:
    POJ1470 Closest Common Ancestors(LCA入门)
    POJ1330 Nearest Common Ancestors(倍增LCA算法求无边权树的模板)
    HDU3078 Network (倍增LCA算法求树链)
    HDU2874 Connections between cities(并查集+倍增LCA算法求森林最短路)
    HDU2586 How far away?(倍增LCA算法求带边权树上最短路)
    POJ1062 昂贵的聘礼
    HDU4725 The Shortest Path in Nya Graph(堆优化的dijkstra算法)
    数据仓库详解:包括概念、架构及设计
    利用行为标签构建用户画像
    Spark SQL深度理解篇:模块实现、代码结构及执行流程总览(2)
  • 原文地址:https://www.cnblogs.com/lgh1992314/p/5834997.html
Copyright © 2011-2022 走看看