POJ 1258 Agri-Net

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

4
0 4 9 21
4 0 8 17
9 8 0 16
21 17 16 0
Sample Output

28

译：

农民约翰已经当选市长的城市!他的竞选承诺之一就是把网络连接中的所有农场地区。当然,他需要你的帮助。
农民约翰下令高速连接他的农场,去分享他的连接与其他农民。成本降到最低,他想躺最少的光纤连接农场到其他所有的农场。
给定的列表需要多少纤维连接每一对农场,你必须找到所需的最少的纤维连接起来。每个农场都必须连接到其他一些农场,这样一个包可以从农场到其他任何一个农场。
任何两个农场之间的距离不会超过100000。

典型的最小生成树，因为是稠密图，可直接用Prim算法解，一次就A了

#include<cstdio>#include<iostream>using namespace std;int map[110][110];int lc[110];const int inf=999999999;int main () {    int n;    while(cin>>n) {        for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)           cin>>map[i][j];        for(int i=1;i<=n;i++)            lc[i]=map[1][i];        int dis=0;        for(int i=1;i<n;i++) {            int mindis=inf;            int minone;            for(int j=1;j<=n;j++) {                if(lc[j]&&mindis>lc[j]) {                    mindis=lc[j];                    minone=j;                }            }            dis+=lc[minone];            lc[minone]=0;            for(int j=1;j<=n;j++) {                if(lc[j]>map[minone][j])                   lc[j]=map[minone][j];            }        }        cout<<dis<<endl;    }    return 0;}