hdoj_1162Eddy's picture（最小生成树）-优快云博客

本文介绍了一个有趣的编程问题：通过构建最小生成树来计算连接所有绘画坐标点所需的最短线段总长度。文章提供了完整的C++实现代码，并采用Kruskal算法解决该问题。

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Eddy's picture

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4953 Accepted Submission(s): 2448

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

最小生成树，

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
using namespace std;
#pragma warning(disable : 4996)
#define MAX 10000
typedef struct edge
{
	int x, y;
	double w;
}edge;

edge e[MAX];
int father[MAX], ranks[MAX];

bool cmp(edge a,edge b)
{
	return a.w < b.w;
}

void Make_Set(int n)
{
	for(int i = 1; i <= n; i++)
	{
		father[i] = i;
		ranks[i] = 0;
	}
}

int Find_Set(int x)
{
	if(x != father[x])
		father[x] = Find_Set(father[x]);
	return father[x];
}

void Merge_Set(int x, int y)
{
	x = Find_Set(x);
	y = Find_Set(y);
	if(x == y) return;
	if(ranks[x] > ranks[y])
	{
		father[y] = x;
	}
	else if(ranks[x] < ranks[y])
	{
		father[x] = y;
	}
	else 
	{
		ranks[y]++;
		father[x] = y;
	}
}

int main()
{
	freopen("in.txt", "r", stdin);
	double point[MAX][2] = {0.0};
	int n, count;
	double len;
	while (scanf("%d", &n) != EOF)
	{
		count = 0;
		for(int i = 1; i <= n; i++)
		{
			scanf("%lf %lf", &point[i][0], &point[i][1]);
		}
		for(int i = 1; i <= n; i++)
		{
			for(int j = i + 1; j <= n; j++)
			{
				len = sqrt((point[i][0] - point[j][0]) * (point[i][0] - point[j][0]) + (point[i][1] - point[j][1]) * (point[i][1] - point[j][1]));
				if(len > 0)
				{
					e[count].x = i;
					e[count].y = j;
					e[count++].w = len;
				}
			}
		}
		Make_Set(count - 1);
		sort(e, e + count , cmp);
		/*for(int i = 0; i < count; i++)
		{
			cout << e[i].x << " " << e[i].y << " " << e[i].w << endl;
		}*/
		double sum = 0.0;
		for (int i = 0; i < count; i++)
		{
			int x = Find_Set(e[i].x);
			int y = Find_Set(e[i].y);
			if(x != y)
			{
				Merge_Set(e[i].x, e[i].y);
				sum += e[i].w;
			}
		}
		printf("%.2lf\n", sum);
	}
	return 0;
}

转载于:https://www.cnblogs.com/lgh1992314/archive/2013/05/09/5835071.html