poj2395 Out of Hay (prim算法)
来源:互联网 发布:勇者之塔守护进阶数据 编辑:程序博客网 时间:2024/05/17 09:09
Out of Hay
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14516 Accepted: 5645
Description
The cows have run out of hay, a horrible event that must be remedied immediately. Bessie intends to visit the other farms to survey their hay situation. There are N (2 <= N <= 2,000) farms (numbered 1..N); Bessie starts at Farm 1. She’ll traverse some or all of the M (1 <= M <= 10,000) two-way roads whose length does not exceed 1,000,000,000 that connect the farms. Some farms may be multiply connected with different length roads. All farms are connected one way or another to Farm 1.
Bessie is trying to decide how large a waterskin she will need. She knows that she needs one ounce of water for each unit of length of a road. Since she can get more water at each farm, she’s only concerned about the length of the longest road. Of course, she plans her route between farms such that she minimizes the amount of water she must carry.
Help Bessie know the largest amount of water she will ever have to carry: what is the length of longest road she’ll have to travel between any two farms, presuming she chooses routes that minimize that number? This means, of course, that she might backtrack over a road in order to minimize the length of the longest road she’ll have to traverse.
Input
Line 1: Two space-separated integers, N and M.
Lines 2..1+M: Line i+1 contains three space-separated integers, A_i, B_i, and L_i, describing a road from A_i to B_i of length L_i.
OutputLine 1: A single integer that is the length of the longest road required to be traversed.
Sample Input
3 3
1 2 23
2 3 1000
1 3 43
Sample Output
43
Hint
OUTPUT DETAILS:
In order to reach farm 2, Bessie travels along a road of length 23. To reach farm 3, Bessie travels along a road of length 43. With capacity 43, she can travel along these roads provided that she refills her tank to maximum capacity before she starts down a road.
Source
USACO 2005 March Silver
本道题一次通过,只要之前练习过最小生成树prim算法的题,我想很容易通过的,直接上代码了。
#include<iostream>#include<cstring>using namespace std;#define infi 0x7fffffff;typedef struct graph{ int verNum,arcNum; int adjMatrix[2005][2005];}graph;int visited[2005];int closeEdge[2005];int prim(graph *G){ int ans=-1; memset(visited,0,sizeof(visited)); visited[0]=1; closeEdge[0]=0; for(int i=1;i<=G->verNum;i++) { closeEdge[i]=G->adjMatrix[i][0]; } for(int j=1;j<G->verNum;j++) { int minc=infi; int p=-1; for(int i=0;i<G->verNum;i++) { if(visited[i]==0 && closeEdge[i]<minc) { minc=closeEdge[i]; p=i; } } // cout<<minc<<endl; visited[p]=1; closeEdge[p]=0; if(ans<minc) { ans=minc; } for(int i=0;i<G->verNum;i++) { if(visited[i]==0 && closeEdge[i]>G->adjMatrix[p][i]) { closeEdge[i]=G->adjMatrix[p][i]; } } } return ans;}int main(){ int n,m; while(cin>>n>>m) { graph *G=new graph; G->verNum=n; G->arcNum=m; for(int i=0;i<G->verNum;i++) for(int j=0;j<G->verNum;j++) { if(i==j) G->adjMatrix[i][j]=0; else G->adjMatrix[i][j]=0x7fffffff; } for(int i=0;i<G->arcNum;i++) { int a,b,c; cin>>a>>b>>c; if(G->adjMatrix[a-1][b-1]>c) { G->adjMatrix[a-1][b-1]=c; G->adjMatrix[b-1][a-1]=c; } } cout<<prim(G)<<endl; } return 0;}
- poj2395 Out of Hay (prim算法)
- POJ2395--Out of Hay
- POJ2395---Out of Hay
- POJ2395--Out of Hay
- POJ2395-Out of Hay
- POJ2395 Out of Hay
- POJ2395 Out of Hay
- POJ2395 Out of Hay 【Dijkstra】
- POJ2395 Out of Hay【Kruskal】
- Out of Hay(poj2395)(并查集)
- POJ2395 Out of Hay 最小生成树
- POJ2395 Out of Hay 最小生成树
- POJ2395 Out of Hay(最小生成树)
- poj2395 Out of Hay(最小生成树)
- poj2395--Out of Hay(最小生成树)
- poj2395 Out of Hay , 求MST的最长边
- TOJ 1638 Out of Hay / prim
- POJ 2395 Out of Hay(最小生成树—prim算法记录最大边)
- 解决ScrollView和GridView滑动冲突的问题
- C++很不错的网址
- 面试基础整理(一)---Java中==和equals的区别
- 如何搭建网站
- 远程访问jupyter notebook
- poj2395 Out of Hay (prim算法)
- charset
- 自定义seekbar中,圆球显示不全被覆盖掉一部分问题
- linux socket网络编程详解
- 面试经典题Handler机制回答
- MySQL基本操作SQL语句
- vlc代码分析(3)——输入模块
- 深入java之java语言基础(一)
- js表单验证方法1