专题四 Problem B

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

3 1.0 1.0 2.0 2.0 2.0 4.0

 

Sample Output

3.41

思路：

这题还是一条最小生成树问题，给出N个定点，求出N个点的最段路径，先求出图的生成树的邻接矩阵，然后用prim算法遍历。

ac代码：

#include
#include
#include
#include
#include
using namespace std;

struct Node
{
    double x;
    double y;
}A[110];

double G[110][110],low[110];
int vis[110];

void Prim(int N)
{
    memset(vis,0,sizeof(vis));
    int pos = 1;
    vis[pos] = 1;
    double ans = 0;
    for(int i = 1; i <= N; i++)
        if(i != pos)
            low[i] = G[pos][i];

    for(int i = 1; i < N; i++)
    {
        double Min = 0x7ffffff;
        for(int j = 1; j <= N; j++)
        {
            if(!vis[j] && Min > low[j])
            {
                Min = low[j];
                pos = j;
            }
        }
        vis[pos] = 1;
        ans += Min;
        for(int j = 1; j <= N; j++)
        {
            if(!vis[j] && low[j] > G[pos][j])
                low[j] = G[pos][j];
        }
    }
    printf("%.2lf\n",ans);
}

int main()
{
    int N;
    double Dist,x,y;
    while(cin >> N)
    {
        for(int i = 1; i <= N; i++)
        {
            for(int j = 1; j <= N; j++)
                G[i][j] = 0x7fffff;
        }
        memset(A,0,sizeof(A));
        for(int i = 1; i <= N; i++)
        {
            cin >> A[i].x >> A[i].y;
        }
        for(int i = 1; i <= N; i++)
        {
            for(int j = 1; j <= N; j++)
            {
                x = A[i].x - A[j].x;
                y = A[i].y - A[j].y;
                Dist = sqrt(x*x+y*y);
                G[i][j] = Dist;
            }
        }
        Prim(N);
    }

    return 0;
}