Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 48633 | Accepted: 20146 |
Description
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
Output
Sample Input
4 0 4 9 21 4 0 8 17 9 8 0 16 21 17 16 0
Sample Output
28
題目:這就是一道裸的Prim的題:
代碼如下:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define MAXN 500
using namespace std;
int n,sum;
int dis[MAXN][MAXN];
int cost[MAXN];
bool vis[MAXN];
void Prim(int start)
{
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
cost[i]=dis[start][i];
sum=0;
vis[start]=true;
for(int i=2;i<=n;i++)
{
int nm=0x3f3f3f,k;
for(int j=1;j<=n;j++)
{
if(!vis[j] && cost[j]<nm)
{
nm=cost[j];
k=j;
}
}
vis[k]=true;
sum+=nm;
for(int j=1;j<=n;j++)
{
if(!vis[j] && cost[j]>dis[k][j])
{
cost[j]=dis[k][j];
}
}
}
}
int main()
{
memset(dis,0x3f3f3f,sizeof(dis));
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
scanf("%d",&dis[i][j]);
}
Prim(1);
printf("%d\n",sum);
}
return 0;
}