hdu 1102 Constructing Roads （Prim算法）

Posted 2020-09-02 风中的簌雨

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1102

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21947    Accepted Submission(s): 8448

Problem Description

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

 

 

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

 

 

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

 

 

Sample Input

3

0 990 692

990 0 179

692 179 0

1

1 2

 

Sample Output

179

 

题目大意：有N个村庄，现在需要修一些路，使得任意两个村庄之间可以连通（可以间接相连），给出每条路的花费，求出最少需要的花费。已知一些路已经修好了。

 

解题思路：根据题目的输入数据，可以选用以顶点为主导的Prim算法求MST。

 

AC代码：

19970675

2017-03-02 20:21:34

Accepted

1102

15MS

1456K

878 B

G++

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define inf 0x3f3f3f3f
 4 
 5 int map[110][110],vis[110],dis[110];
 6 int n;
 7 
 8 void prim()
 9 {
10     int i,j,pos,min,sum = 0;
11     for (i = 1; i <= n; i ++)
12     {
13         vis[i] = 0;
14         dis[i] = map[1][i];
15     }
16     vis[1] = 1; dis[1] = 0;
17     for (i = 1; i < n; i ++)
18     {
19         pos = -1; min = inf;
20         for (j = 1; j <= n; j ++)
21         {
22             if (!vis[j] && min > dis[j])
23             {
24                 min = dis[j];
25                 pos = j;
26             }
27         }
28         vis[pos] = 1;
29         sum += min;
30         for (j = 1; j <= n; j ++)
31             if (!vis[j] && dis[j] > map[pos][j])
32                 dis[j] = map[pos][j];
33     }
34     printf("%d\\n",sum);
35 }
36 int main ()
37 {
38     int i,j,m,a,b;
39     while (~scanf("%d",&n))
40     {
41         for (i = 1; i <= n; i ++)
42         for (j = 1; j <= n; j ++)
43             scanf("%d",&map[i][j]);
44         scanf("%d",&m);
45         for (i = 0; i < m; i ++)
46         {
47             scanf("%d%d",&a,&b);
48             map[a][b] = map[b][a] = 0;
49         }
50         prim();
51     }
52     return 0;
53 }

 

算法理解： http://www.cnblogs.com/yoke/p/6506492.html