最小生成树 prim

 Agri-Net

 POJ - 1258 

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

 

题意 : 简单来说 给n个点,给出 每两点之间的距离,然你算出来所有点之间最短的距离的和.

思路:小菜鸡第一次做最小生成树, 用 prim算法 跟迪杰斯特拉很想,唯一不同的是 更新的时候更新的是两点之间最短的距离,而不是从源点到该点的距离,每次找的都是最近的边存进dis数组里,然后加和,很像迪杰斯特拉,代码没什么差距,理解就好....

代码:

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
typedef long long ll;
const ll INF = 1000000000;
int n;
int d[10005];
int vis[10005];
int maze[105][105];

ll prim(){
    memset(d,INF,sizeof(d));
    memset(vis,0,sizeof(vis));
    for(int i = 1; i <= n; i++){
        d[i] = maze[1][i];//初始化
    }
    vis[1] = 1;
    int flag;
    ll sum = 0;
    for(int i = 2; i <= n; i ++){
        ll maxn = INF;
        for(int j = 1; j<= n;j++){
            if(!vis[j]&& maxn > d[j]){//找到 最短的边以及点.
                maxn = d[j];
                flag = j;
            }
        }
        vis[flag] = 1;
        sum += maxn;//将最短的距离求和
        for(int j = 1; j<= n; j++){
            if(!vis[j]){
                d[j] = min(d[j],maze[flag][j]);//更新该点与其相连的点之间的距离,只是该点与与它相连的点之间的距离而不是源点.
            }
        }

    }

    return sum;
}
int main()
{
    while(~scanf("%d",&n)){
        for(int i = 1; i <= n;i ++){
            for(int j = 1; j<= n;j ++){
                scanf("%d",&maze[i][j]);
            }
        }
        ll ans = prim();
        printf("%lld\n",ans);
    }
}