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Agri-Net
Time Limit: 1000MS Memory limit: 10000K
题目描写叙述
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.
The distance between any two farms will not exceed 100,000.
输入
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.
输出
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.
演示样例输入
4 0 4 9 21 4 0 8 17 9 8 0 16 21 17 16 0
演示样例输出
28
#include <stdio.h>
#include <string.h>
int Map[110][110];
int eveco[110], vis[110];
int n,m;
int Prim(int n)
{
int cost = 0;
memset(vis, false, sizeof(vis));
for(int i=1; i<=n; i++)
eveco[i] = Map[1][i];
vis[1] = true;
for(int i=1; i<n; i++)
{
int m = 0x3f3f3f3f;
int pos;
for(int j=1; j<=n; j++)
if(!vis[j] && eveco[j] < m)
m = eveco[j], pos = j;
if(m == 0x3f3f3f3f)
break;
cost += m;
vis[pos] = true;
for(int j=1; j<=n; j++)
if(!vis[j] && eveco[j] > Map[pos][j])
eveco[j] = Map[pos][j];
}
return cost;
}
int main()
{
while(~scanf("%d",&n))
{
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
scanf("%d",&Map[i][j]);
printf("%d\n",Prim(n));
}
return 0;
}
版权声明：本文为博主原创文章，未经博主同意不得转载。
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