POJ - 2421 Constructing Roads（最小生成树 prim算法）

Constructing Roads

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 25971		Accepted: 11367

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

Code

#include 
#include 
#include 
#include 
#define N 110
#define INF 0x3f3f3f3f

using namespace std;

int map[N][N];
int vis[N];
int dis[N];
int ans,n;

void prim()
{
    int i,j,min,pos;
    for(i=1; i<=n; i++)
    {
        dis[i] = map[1][i];
        vis[i] = 0;
    }
    vis[1] = 1;
    for(i=1; i<=n-1; i++)
    {
        min = INF;
        for(j=1; j<=n; j++)
        {
            if(!vis[j] && dis[j] < min)
            {
                min = dis[j];
                pos = j;
            }
        }
        ans += min;
        vis[pos] = 1;
        for(j=1; j<=n; j++)
        {
            if(!vis[j] && dis[j] > map[pos][j])
            {
                dis[j] = map[pos][j];
            }
        }
    }
}

int main()
{
    int m,i,j,a,b;
    while(~scanf("%d",&n))
    {
        ans = 0;
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=n; j++)
            {
                scanf("%d",&map[i][j]);
            }
        }
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d %d",&a,&b);
            map[a][b] = map[b][a] = 0;
        }
        prim();
        printf("%d\n",ans);
    }
    return 0;
}

反思：

题目大意：多组输入，给出图的点数n和图的邻接矩阵，接下来有m组数据，每组包含a和b两个整数，说明a和b两个村庄连通。最小生成树模板题，因为题目给出了邻接矩阵，所以连图的初始化都省了♪(ﾟ▽^*)ﾉ⌒☆。已经连通的两个村庄权值置为0，然后用prim即可。