Program4_A

 我现在做的是第四专题编号为1001的试题，具体内容如下所示：

Problem A

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 24   Accepted Submission(s) : 16
[align=left]Problem Description[/align]
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. <br><br>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.<br>
 

[align=left]Input[/align]
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.<br><br>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.<br>
 

[align=left]Output[/align]
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. <br>
 

[align=left]Sample Input[/align]

3
0 990 692
990 0 179
692 179 0
1
1 2

 

[align=left]Sample Output[/align]

179

简单题意：

在修建之前有的村庄之间有已经建造的道路，求使所有村庄连在一起需要再建造的道路的最小值

解题思路：

最小生成树的题目，主要是并查集的运用，将关系矩阵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....

编写代码：

#include<stdio.h>

#include<string.h>

#include<algorithm>

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)

    {

        per[i]=i;

    }

}

int find(int x)

{

    int r=x;

    while(r!=per[r])

    {

        r=per[r];

    }

    int i=x,j;

    while(i!=r)

    {

        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)

    {

        per[fy]=fx;

        ++cnt;

        kase=1;

    }

}

int cmp(road a,road b)

{

    return a.len<b.len;

}

int main()

{

    int a,b,i,j,c,sum;

    while(~scanf("%d",&n))

    {

        init();

        cnt=c=0;

        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);

        for(i=0;cnt<n-1;++i)

        {

            kase=0;

            join(x[i].a,x[i].b);

            if(kase)

            {

                sum+=x[i].len;

            }

        }

        printf("%d\n",sum);

    }

    return 0;

}