Hdu 1102 Constructing Roads【最小生成树】

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 17176    Accepted Submission(s): 6524

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

 

最小生成树的题目，自己用的克鲁斯克尔算法做的，主要是并查集的运用，刚见到这个题自己想复杂了，其实还是比较简单的，虽然给出的是关系矩阵，但是很简单就可以转化为经典的顶点和边的模型，而且长短已知，只需要判断什么时候统计和累加就可以了，因为已经给出了某些已知的路径，那么就先把这些路径涉及的端点先合并，并且在合并的时候统计有效的道路数量（不形成环），然后下面就是遍历查找，直到 n-1 （n个端点）条边全部找到，最小生成树已经生成，输出相应的累加数值....

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
int per[10005],n,kase,cnt;
struct road
{
    int a,b;
    int len;
}x[10005];
void init()
{
    for(int i=1;i<=n;++i)//初始化 
    {
        per[i]=i;
    }
}
int find(int x)
{
    int r=x;
    while(r!=per[r])//找根节点 
    {
        r=per[r];
    }
    int i=x,j;
    while(i!=r)//压缩路径 
    {
        j=per[i];per[i]=r;i=j;
    }
    return r;
}
void join(int x,int y)
{
    int fx=find(x),fy=find(y);
    if(fx!=fy)//可以合并 
    {
        per[fy]=fx;//合并 
        ++cnt;kase=1;//边数+1，标记状态 
    }
}
int cmp(road a,road b)
{
    return a.len<b.len;//距离由小到大排序 
}
int main()
{
    int a,b,i,j,c,sum;
    while(~scanf("%d",&n))
    {
        init();cnt=c=0;//初始化 
        for(i=0;i<n;++i)
        {
        	for(j=0;j<n;++j)
        	{
        		scanf("%d",&a);
        		if(i!=j)
        		{
        			x[c].a=i+1;x[c].b=j+1;
        			x[c].len=a;
        			++c;
				}
			}
        }
        int m,sum=0;
        scanf("%d",&m);
        for(i=0;i<m;++i)
        {
        	scanf("%d%d",&a,&b);
        	join(a,b);
		}
		sort(x,x+c,cmp);//按路径长短排序
        for(i=0;cnt<n-1;++i)//遍历 
        {
            kase=0;
            join(x[i].a,x[i].b);//否则合并 
            if(kase)//如果成功合并
            {
                sum+=x[i].len;//长度累加 
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}