题目链接：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): 14172    Accepted Submission(s):
5402

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：第一行表示n*n的矩阵，然后输入矩阵，表示的就是1到1的距离，1到2的距离。。。。第二行是2到1的距离，2到2的距离。。。。以此类推~

然后在输入一行，表示下面有几组数据，下面输入的两个数就是表示两个之间有路，不需要再建了，最后就是输出能够连通所有道路的最短路了！！！

一个最小生成树prim就可以解决了！哇哈哈~

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 int map[1010][1010],node[1010],Min,n,a,b;
 5 const int INF=999999999;
 6
 7 int prim()
 8 {
 9     int vis[1010]= {0};
10     int tm=1,m,s=0;
11     vis[tm]=1;
12     node[tm]=0;
13     for (int k=1; k<=n; k++)
14     {
15         Min=INF;
16         for (int i=1; i<=n; i++)
17             if (!vis[i])
18             {
19                 if (node[i]>map[tm][i])
20                     node[i]=map[tm][i];
21                 if (Min>node[i])
22                 {
23                     Min=node[i];
24                     m=i;
25                 }
26                 //s+=Min;
27             }
28         tm=m;
29         vis[m]=1;
30
31     }
32     for (int i=1; i<=n; i++)
33         s+=node[i];
34     return s;
35 }
36
37 int main ()
38 {
39     while (cin>>n)
40     {
41         //cout<<n<<endl;
42         for (int i=1; i<=n; i++)
43         {
44             node[i]=INF;
45             for (int j=1; j<=n; j++)
46                 map[i][j]=INF;
47         }
48         for (int i=1; i<=n; i++)
49         {
50             for (int j=1; j<=n; j++)
51             {
52                 cin>>map[i][j];
53             }
54         }
55         int q;
56         cin>>q;
57         while (q--)
58         {
59             cin>>a>>b;
60             map[a][b]=map[b][a]=0;
61         }
62         printf ("%d\n",prim());
63     }
64     return 0;
65 }