标签：des style blog color os strong
Agri-Net
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 36780 |
|
Accepted: 14825 |
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
单纯的求最小生成树普林斯算法，要说是水题，却不是菜鸟都能轻松过的。且学且刷~~
#include<iostream>
#include<cstring>
using namespace std;
int a[101][101],d[101],used[101],n;
int main(){
while(cin>>n&&n>=3&&n<=100){
memset(d,0,sizeof(d));
memset(a,0,sizeof(a));
memset(used,0,sizeof(used));
int sum=0;
used[0]=1;
d[0]=999999; //尽可能大，不要错写成0
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
cin>>a[i][j];
for(int k=1;k<n;k++)d[k]=a[0][k];//数组d用于动态记录点到最小生成树的距离
for(int i=0;i<n-1;i++){
int flag=0;
for(int j=1;j<n;j++)
if(d[j]<d[flag]&&!used[j]) flag=j;//遍历寻找树外到最小树距离最小的点
used[flag]=1;
sum+=d[flag];
for(int j=1;j<n;j++) //节点flag加入最小生成树，更新数据
if(!used[j])
d[j]=min(d[j],a[j][flag]);
}
cout<<sum<<endl;
}
return 0;
}
poj1258 最小生成树 普林姆,码迷,mamicode.com
poj1258 最小生成树 普林姆
标签：des style blog color os strong
原文地址：http://blog.csdn.net/lindonglian/article/details/24736489
