HDOJ 1162 Eddy's picture 最小生成树

Eddy的画作与最小生成树

最新推荐文章于 2025-12-02 21:05:32 发布

原创最新推荐文章于 2025-12-02 21:05:32 发布 · 239 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#acm #c++ #最小生成树

最小生成树专栏收录该内容

3 篇文章

订阅专栏

Eddy热爱画画并希望改善技艺以获得朋友的认可。为此，他提出了一道编程题：寻找连接所有点所需的最短墨水线总长度。此问题可通过实现Prim算法求解最小生成树来完成。

Eddy's picture

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10754 Accepted Submission(s): 5431

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

Recommend

JGShining | We have carefully selected several similar problems for you: 1301 1232 1875 1879 1272

一道赤裸裸的最小生成树题目，因为任意两点都可以连线，是个稠密图，用prim算法更好。

不过数据范围小，kruskal也可以AC。

我的代码使用prim算法

#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 105;
const int inf = 1e5;
int n,i,j,k;
double x[maxn],y[maxn],a[maxn][maxn],d[maxn];
double dis;
bool used[maxn];


double Distance(int p , int q){
     return sqrt((x[p]-x[q])*(x[p]-x[q]) + (y[p]-y[q])*(y[p]-y[q]));
}

void input(){
   for (i=1; i<=n; i++) scanf("%lf %lf",&x[i],&y[i]);
}

void init(){
    dis = 0;
    for (i=1; i<n; i++)
      for (j=i+1; j<=n; j++) {
         a[j][i] = a[i][j] = Distance(i,j);
      }
    memset(used,0,sizeof(used));
    for (i=1; i<=n; i++) d[i] = 1.0*inf;
}


void prim(){
    int s=1;
    double min_dis;
    used[s] = 1;
    for (i=1; i<=n; i++) {
        for (j=1; j<=n; j++) d[j] = min (d[j],a[s][j]);
        min_dis = 1.0*inf;
        for (j=1; j<=n; j++) if (!used[j] && d[j]<min_dis) {
            min_dis = d[j];
            s = j;
        }
        used[s] = 1;
        dis += d[s];
    }
}

int main(){
    while (~scanf("%d",&n)){
        input();
        init();
        prim();
        printf("%.2lf\n",dis);
    }
    return 0;
}