HDU 1102 Constructing Roads（Kruskal最小生成树求最小花费）

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): 20248 Accepted Submission(s): 7743

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：
就是总共有n个村庄，然后给你一个矩阵图，表示每两个村庄之间的距离。然后再告诉你目前有多少个村庄是已经联系的，让你求出最少要修多少米路可以使得任意两个村庄之间可以相互到达。

思路构造：
首先任意两个地点之间可以相互到达，并且建造路所花的花费最少，明显这就是一道最小生成树题目。
剩下要做的就是建边，然后求最小生成树，我用的是Kruskal算法。
下面是AC代码：

#include
#include
#include
using namespace std;

struct node
{
    int u,v,cost;
}a[10005];
int pre[105];
int fin(int x)
{
    if(pre[x]==x)
    {
        return x;
    }
    else
    {
        return pre[x]=fin(pre[x]);
    }
}

void join(int x,int y)
{
    int t1=fin(x);
    int t2=fin(y);
    if(t1!=t2)
    {
        pre[t1]=t2;
    }
}

bool cmp(node x,node y)
{
    return x.costint main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int num;
        for(int i=1;i<=n;i++)
        {
            pre[i]=i;
        }
        int sum=0;
        for(int i=1;i<=n;i++)//建边
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&num);
                if(i!=j)
                {
                   a[sum].u=i;
                   a[sum].v=j;
                   a[sum++].cost=num;
                }
            }
        }
        int Q,c,d,s=0;//s是优化，因为最小生成树中共有n-1条边
        scanf("%d",&Q);
        for(int i=0;i//建树
        {
            scanf("%d%d",&c,&d);
            if(fin(c)!=fin(d))
            {
                join(c,d);
                s++;
            }
        }
        int re=0;
        sort(a,a+sum,cmp);
        for(int i=0;iif(fin(a[i].u)!=fin(a[i].v))
            {
                join(a[i].u,a[i].v);
                re+=a[i].cost;
                s++;
            }
            if(s==n-1)
            {
                break;
            }
        }
        printf("%d\n",re);
    }
    return 0;
}