POJ-1258 最小生成树 prim算法

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我是靠谱客的博主友好铃铛，最近开发中收集的这篇文章主要介绍POJ-1258 最小生成树 prim算法，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

prim（假设图是连通的，转化成连通网的最小生成树）

n个顶点----全部连通----n-1条路

mapp记录所有边

dis 记录权值

vis 标记是否经过

搜索最小边

把最小边加进去

累加到ans里

最后记得更新权值

Agri-Net

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 39435		Accepted: 15987

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

Sample Output

最小生成树的模板题，不是很难，但是小循环的循环变量跟大循环用成一个了。。

改了好久才发现这里错了。。而且，坐标的下标标错了。好桑心。。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int mapp[505][505];
int dis[5050];
int vis[5050];
int ans,flag;
int prim(int m)
{
int i,j;
int pos;
int minn;
memset(vis,0,sizeof(vis));
// memset(dis,0,sizeof(dis));
ans = 0;
for(i=1; i<=m; i++)
{
dis[i] = mapp[1][i];
}
vis[1] = 1;
for(i=1; i<=m-1; i++)
{
minn = INF;
for(j=1; j<=m; j++)
{
if(vis[j]==0&&minn>dis[j])
{
pos = j;
minn = dis[j];
}
}
vis[pos] = 1;
ans = ans + minn;
if(minn>=INF)
{
break;
}
for(j=1; j<=m; j++)
{
if(vis[j]==0&&dis[j]>mapp[pos][j])
{
dis[j] = mapp[pos][j];
}
}
}
return ans;
}
int main()
{
int n;
int a;
int t;
while(scanf("%d",&n)!=EOF)
{
int i,j;
for(i=1; i<=n; i++)
{
for(j=1; j<=n; j++)
{
scanf("%d",&a);
mapp[i][j] = a;
mapp[j][i] = a;
}
}
t = prim(n);
printf("%dn",t);
}
return 0;
}