poj1258解題報告

Agri-Net

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15626		Accepted: 6327

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.

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

40 4 9 214 0 8 179 8 0 1621 17 16 0

Sample Output

28

題目大意:給定一個N村莊和一個N^2的鄰接矩陣表示每兩個村莊間的距離，求將這N個點聯通所需的最小電線長度  。。

思路：典型的最小產生樹,我用的prim演算法，就是開始隨便取一個點，找出這個點和為加入的哪個點距離最短，再將那個點也加入到被選中的

集合中，反覆重複這個過程直到將所有點加入到被選中集合。。。47ms水鍋

#include<iostream>using namespace std;int main(){int dis[101][101],lock[101],n,i,j,min,finded,p,s;while(cin>>n&&n){memset(lock,0,sizeof(lock));for(i=0;i<n;i++)for(j=0;j<n;j++)cin>>dis[i][j];lock[0]=1;finded=1;s=0;while(finded!=n){min=100000000;for(i=0;i<n;i++)for(j=0;j<n;j++)if(lock[i]==1&&lock[j]==0)if(dis[i][j]<min){min=dis[i][j];p=j;}s+=min;lock[p]=1;finded++;}cout<<s<<endl;}return 0;}