首页 > 代码库 > poj 1258
poj 1258
Agri-Net
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 38805 | Accepted: 15674 |
Description
Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.
Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.
Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.
The distance between any two farms will not exceed 100,000.
Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.
Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.
The distance between any two farms will not exceed 100,000.
Input
The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.
Output
For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.
Sample Input
4 0 4 9 21 4 0 8 17 9 8 0 16 21 17 16 0
Sample Output
28
Source
USACO 102
直接套prim算法就AC了
#include<iostream> using namespace std; int n; int G[105][105]; void prim(){ int vis[105]; int dis[105]; int sum=0; for(int i=1;i<=n;i++){ vis[i]=0; dis[i]=999999; } dis[1]=0; for(int i=1;i<=n;i++){ int v,min=999999; for(int j=1;j<=n;j++){ if(!vis[j] && min>dis[j]){ min=dis[j]; v=j; } } vis[v]=1; sum+=min; for(int j=1;j<=n;j++){ if(!vis[j] && G[v][j] && G[v][j]<dis[j]) dis[j]=G[v][j]; } } cout<<sum<<endl; } int main(){ while(cin>>n){ if(n<3 || n>100) continue; for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) cin>>G[i][j]; prim(); } return 0; }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。