HDU - 1102 - Constructing Roads （最小生成树--prim算法！！）

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14890    Accepted Submission(s): 5674

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

 

Source

kicc

 

最小生成树 Prime算法：首先有两个集合V{},边的集合E{}。先选择一个点，加入到V集合中，然后选择和这个V集合中的点相关连的边中最小的，将边的另一头的点加入到V集合中，该边加入到E集合中，V集合中的点是一个连通图，每次找和这个连通图相连的最短边，来更新。

AC代码：

#include 
#include 
#include 
using namespace std;

const int N = 105;
const int INF = 0xffffff;
int map[N][N], n;
int vis[N*(N+1)/2];
int dis[N*(N+1)/2];

void prim()
{
    int i, j, t;
    memset(vis, 0, sizeof(vis));
    for(i=1; i<=n; i++)
		dis[i]=map[1][i];
    int loc, sum=0;
    for(i=1; i<=n; i++)
	{
       int min = INF;
       for(j=1; j<=n; j++)
	   {
           if(!vis[j] && dis[j]