hdu1102 Constructing Roads (简单最小生成树Prim算法)

最小生成树算法解析

最新推荐文章于 2020-07-06 09:36:29 发布

转载最新推荐文章于 2020-07-06 09:36:29 发布 · 209 阅读

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原文链接：http://www.cnblogs.com/mis-xiao/p/3931064.html

本文介绍了一个关于构建最小生成树的问题，通过具体的样例输入输出展示了如何使用Prim算法解决村庄间道路连接问题，以实现所有村庄间的最短总路程。

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

0 990 692

990 0 179

692 179 0

1 2

Sample Output

179

很简单的一个最小生成树题意也很简单看完直接AC了好开心O(∩_∩)O~~

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 
 6 using namespace std;
 7 
 8 #define inf 99999999
 9 
10 int map[105][105],path,node[105],vis[105],n;
11 
12 void Prim()
13 {
14     int ans=1;
15     vis[1]=1;
16     node[1]=0;
17     for(int k=1;k<=n;k++)
18     {
19         int minn=inf;
20         for(int i=1;i<=n;i++)
21         if(!vis[i]&&minn>node[i])
22         {
23             minn=node[i];
24             ans=i;
25         }
26         vis[ans]=1;
27         path+=node[ans];
28 
29         for(int i=1;i<=n;i++)
30         {
31             if(!vis[i]&&node[i]>map[ans][i])
32             node[i]=map[ans][i];
33         }
34     }
35 }
36 
37 int main()
38 {
39     while(cin>>n)
40     {
41         for(int i=1;i<=n;i++)
42         {
43             node[i]=inf;vis[i]=0;
44             for(int j=1;j<=n;j++)
45             cin>>map[i][j];
46         }
47         int m;
48         cin>>m;
49         while(m--)
50         {
51             int a,b;
52             cin>>a>>b;
53             map[a][b]=map[b][a]=0;
54         }
55         path=0;
56         Prim();
57         cout<<path<<endl;
58     }
59     return 0;
60 }

转载于:https://www.cnblogs.com/mis-xiao/p/3931064.html