HDU 3549 Flow Problem
来源:互联网 发布:淡斑 知乎 编辑:程序博客网 时间:2024/04/28 02:31
Flow Problem
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 7865 Accepted Submission(s): 3659
Problem Description
Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.
Input
The first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)
Output
For each test cases, you should output the maximum flow from source 1 to sink N.
Sample Input
23 21 2 12 3 13 31 2 12 3 11 3 1
Sample Output
Case 1: 1Case 2: 2题意很简单 ,就是一个简单的最大流问题,模板直接过。。。作为一个新手刚学 表示理解的还是不怎么深刻,跟龙哥调了一上午也没发现错在哪。。啥都不说了,都是泪啊 上代码#include <iostream>#include <queue>#include <string.h>#include <stdio.h>#include <stdlib.h>#define N 25#define MAX 9999999using namespace std;int n,m;int vis[N];int map1[N][N],pre[N];int bfs(int s,int e){ memset(vis, 0, sizeof(vis)); memset(pre,-1,sizeof(pre)); pre[s]=0; vis[s]=MAX; queue<int>q; q.push(s); while(!q.empty()) { int k = q.front(); q.pop(); if(k==e) break; for(int i=1; i<=n; i++) { if(map1[k][i]>0 && pre[i]==-1) { pre[i] = k; vis[i] = min(map1[k][i],vis[k]); //一开始while外了,也没错就是交不上,后来放着居然就A了 q.push(i); } } } if(pre[e]==-1) return -1; return vis[e];}int zuida(int s,int e){ int ans=0; while(1) { int y = bfs(s,e); if(y == -1) break; for(int i = n; i != 1; i = pre[i]) { map1[pre[i]][i] -= y; map1[i][pre[i]] += y; } ans+=y; } return ans;}int main(){ int T; int Case = 1; scanf("%d",&T); while(T--) { memset(map1,0,sizeof(map1)); scanf("%d%d",&n,&m); for(int i=1; i<=m; i++) { int a,b,c; scanf("%d%d%d",&a,&b,&c); map1[a][b]+= c; } int ans=zuida(1, n); cout<<"Case "<<Case++<<": "; printf("%d\n",ans); } return 0;}
0 0
- HDU 3549 Flow Problem
- hdu 3549 Flow Problem
- hdu 3549 Flow Problem
- hdu 3549 Flow Problem
- HDU 3549 Flow Problem
- hdu(3549)Flow Problem
- HDU 3549 Flow Problem
- hdu 3549 Flow Problem
- HDU 3549 Flow Problem
- HDU 3549 Flow Problem
- HDU 3549 Flow Problem
- HDU 3549 Flow Problem
- HDU 3549 Flow Problem
- HDU 3549 Flow Problem
- hdu 3549 Flow Problem
- hdu 3549Flow Problem
- HDU 3549 Flow Problem
- hdu 3549 Flow Problem
- 背包问题的递归解法
- Freemodbus RTU在stm32上的移植分析
- HDU 3729 I'm Telling the Truth(二分图最大匹配)
- 本地js页面
- JAXB基本类型对应关系
- HDU 3549 Flow Problem
- hdoj 1022 Train Problem I(用栈解决法)
- Linux mint 下安装Fcitx中文输入法
- POJ3249 Test for Job 【DAG】+【记忆化搜索】
- 图片加水印工具类
- hdu3549 Flow Problem
- 秒杀多线程第十篇 生产者消费者问题
- 用WebCollector做Web挖掘(实例1)
- google 2015最新的校招测试题与思路