HDU 1162 Eddy's picture MST（基础）

最新推荐文章于 2018-07-01 17:27:00 发布

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本文详细介绍了最小生成树的概念、算法原理及其在实际问题中的应用案例，通过实例演示了如何利用最小生成树解决连接所有点的最短路径问题。

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Eddy's picture

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5539 Accepted Submission(s): 2776

Problem Description

Eddy begins to like painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the result it can be imagined, the friends are not interested in his picture.Eddy feels very puzzled,in order to change all friends 's view to his technical of painting pictures ,so Eddy creates a problem for the his friends of you.
Problem descriptions as follows: Given you some coordinates pionts on a drawing paper, every point links with the ink with the straight line, causes all points finally to link in the same place. How many distants does your duty discover the shortest length which the ink draws?

Input

The first line contains 0 < n <= 100, the number of point. For each point, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the point.

Input contains multiple test cases. Process to the end of file.

Output

Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the points.

Sample Input

Sample Output

  
   3.41

Author

eddy

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最小生成树简单题。

#include<stdio.h>
#include<string.h>
#include<math.h>
#define inf 0x3f3f3f
double g[107][107];
double x[107],y[107],pri[107];
bool vis[107];
int n;

void prime()
{
    int id;
    double minn,count=0;
    memset(vis,false,sizeof(vis));
    for(int i=1; i<=n; i++)
        pri[i]=g[1][i];
    vis[1]=true;
    for(int i=2; i<=n; i++)
    {
        minn=inf;
        for(int j=2; j<=n; j++)
            if(minn>pri[j]&&!vis[j])
            {
                minn=pri[j];
                id=j;
            }
        vis[id]=true;
        count+=minn;
        for(int j=2; j<=n; j++)
            if(!vis[j]&&pri[j]>g[j][id])
                pri[j]=g[j][id];
    }
    printf("%.2f\n",count);
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1; i<=n; i++)
            scanf("%lf%lf",&x[i],&y[i]);
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                double tmp_x,tmp_y,tmp;
                tmp_x=(x[j]-x[i])*(x[j]-x[i]);
                tmp_y=(y[j]-y[i])*(y[j]-y[i]);
                tmp=sqrt(tmp_x+tmp_y);
                g[i][j]=g[j][i]=tmp;
                //printf("aaaffh%lf\n",tmp);
            }
        prime();
    }

    return 0;
}