题目传送门

解题思路:

一道裸的最小生成树,题解传送门

AC代码:

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 
 7 int n,g[101][101],dis[101],ans; 
 8 bool p[101];
 9 
10 void Prim() {
11     dis[1] = 0;
12     p[1] = true;
13     for(int i = 1;i < n; i++) {
14         int k = 0;
15         for(int j = 1;j <= n; j++) 
16             if(!p[j] && dis[j] < dis[k])
17                 k = j;
18         p[k] = true;
19         ans += dis[k];
20         for(int j = 1;j <= n; j++)
21             dis[j] = min(dis[j],g[k][j]);
22     }
23 }
24 
25 int main()
26 {
27     memset(g,0x3f,sizeof(g));
28     scanf("%d",&n);
29     for(int i = 1;i <= n; i++) 
30         for(int j = 1;j <= n; j++)
31             scanf("%d",&g[i][j]);
32     for(int i = 0;i <= n; i++)
33         dis[i] = g[1][i];
34     Prim();
35     printf("%d",ans);
36     return 0;
37 }

标签：include,洛谷,int,题解,P1546,传送门,Agri,101,dis
来源： https://www.cnblogs.com/lipeiyi520/p/11286127.html