POJ-1258 Agri-Net(最小生成树)

Agri-Net
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 51528   Accepted: 21486

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

4
0 4 9 21
4 0 8 17
9 8 0 16
21 17 16 0

Sample Output

28

Source

 

题目大意：FarmerJohn的农村划分为n个区域，需要架设光纤网络连通。每两个区域间架设光纤所需费用不同，给出两两之间的架设价格，求连通所有地块的最小费用。

翻文件夹时偶然发现了这题，标注是最小生成树基本练习，受zhx大神的影响，我是不会去写prim的，于是我就用kruskal水掉了这题，不过一开始被多组数据坑了……

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <queue>
 6 #include <stack>
 7 #include <vector>
 8 #include <iostream>
 9 #include "algorithm"
10 using namespace std;
11 typedef long long LL;
12 const int MAX=105;
13 int n,m;
14 int f[MAX][MAX];
15 int ans;
16 struct Edge{
17     int u,v;
18     int w;
19 }edge[MAX*MAX];
20 int fa[MAX*MAX];
21 int getfather(int x){
22     if (fa[x]==x)
23      return x;
24     return fa[x]=getfather(fa[x]);
25 }
26 bool cmp(Edge x,Edge y){
27     return x.w<y.w;
28 }
29 void init(){
30     int i,j;
31     for (i=1;i<=n;i++)
32      for (j=1;j<=n;j++)
33       scanf("%d",&f[i][j]);
34     m=0;
35     for (i=1;i<=n;i++)
36      for (j=i+1;j<=n;j++)
37      {m++;
38       edge[m].u=i;
39       edge[m].v=j;
40       edge[m].w=f[i][j];
41      }
42     for (i=1;i<=m;i++)
43      fa[i]=i;
44     sort(edge+1,edge+m+1,cmp);
45 }
46 int main(){
47     freopen ("net.in","r",stdin);
48     freopen ("net.out","w",stdout);
49     while (scanf("%d\n",&n)!=EOF){
50     init();int i,j;
51     ans=0;
52     int tx,ty;
53     for (i=1;i<=m;i++)
54     {tx=getfather(edge[i].u);
55      ty=getfather(edge[i].v);
56      if (tx!=ty)
57      {fa[tx]=ty;
58       ans+=edge[i].w;
59      }
60     }
61     printf("%d\n",ans);}
62     return 0;
63 }

 

posted @ 2016-08-01 11:37  lonely_OIer  阅读(120)  评论(0)  编辑  收藏  举报