poj 2421 Constructing Roads 并查集+最小生成树

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

题意及分析：

这个是上篇的那个题（poj2031）的简化。有n个村庄，已存在m条路，现在修一些其它的路，是村庄之间两两联通，并且路的总长度最短。具体过程可以参见上篇题解，就不多说了。

AC代码：

 

#include
#include
#include
#include
#include
#include
#include
#include
#define INF 0x3f3f3f3f
using namespace std;
int root[120],rec[120],vis[120],dis[120],edge[120][120];
int n,num,m;
int fun(int x)
{
    int t=x;

    while(x!=root[x])
        x=root[x];
    while(t!=root[t])
    {
        t=root[t];
        root[t]=x;
    }

    return x;
}
void mer(int a,int b)
{
    int ra=fun(a);
    int rb=fun(b);

    if(ra==rb)
        return;
    root[ra]=rb;
}
int prim()
{
    int i,j,now;

    memset(vis,0,sizeof(vis));
    memset(dis,0x3f,sizeof(dis));
    dis[rec[0]]=0;
    for(i=0;iedge[rec[now]][rec[j]])
                dis[rec[j]]=edge[rec[now]][rec[j]];
    }
    int sum=0;
    for(i=0;iedge[i][j])
                edge[r2][r1]=edge[r1][r2]=edge[i][j];
        }
        if(!num)
            printf("0\n");
        else
            printf("%d\n",prim());
    }

    return 0;
}