题目如下：

Eddy's picture

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7267 Accepted Submission(s): 3676

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 
  

题目大意：

给出n个点的坐标（二维x， y），可以使用直线（无向的）将任意两个点连接起来，求将所有点连接起来形成一个整体（使任何两点之间可达），线的最小距离。

一个简单的最小生成树问题，这里边的权要根据输入的点的坐标信息求出，在输入的时候存储点的数组中下标可以认为是该点的编号，通过O（n^2)的复杂度来求任意两个不同点之间的距离作为权，边的两个顶点就是这段距离两端点的编号。则接下来就是简单的最小生成树算法了，可以使用kruskal也可以使用prim。

kruskal算法实现

#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

const int N = 105;

struct Edge
{
	int x, y;
	double w;
};

struct Point
{
	double x;
	double y;
};

int pre[N];
Point point[N];
Edge edges[N * N / 2];

int i_p, i_e, cnt;
double res;

int root(int x)
{
	if (x != pre[x])
	{
		pre[x] = root(pre[x]);
	}
	return pre[x];
}

bool merge(int x, int y)
{
	int fx = root(x);
	int fy = root(y);

	bool ret = false;

	if (fx != fy)
	{
		pre[fx] = pre[fy];
		ret = true;
		--cnt;
	}
	return ret;
}

void init(int n)
{
	cnt = n;
	res = 0;

	for (int i = 0; i <= n; ++i)
	{
		pre[i] = i;
	}
}

bool cmp(const Edge &a, const Edge &b)
{
	return a.w < b.w;
}

int main()
{
	int n;
	double dx, dy;

	while (scanf("%d", &n) != EOF)
	{
		init(n);

		i_e = i_p = 0;

		for (int i = 0; i < n; ++i)
		{
			scanf("%lf %lf", &dx, &dy);
			point[i_p].x = dx;
			point[i_p].y = dy;
			++i_p;
		}

		for (int i = 0; i < n; ++i)
		{
			for (int j = i + 1; j < n; ++j)
			{
				edges[i_e].x = i;
				edges[i_e].y = j;
				double dd = (point[i].x - point[j].x) * (point[i].x - point[j].x);
				dd += (point[i].y - point[j].y) * (point[i].y - point[j].y);
				edges[i_e].w = sqrt(dd);
				++i_e;
			}
		}

		sort(edges, edges + i_e, cmp);

		//the cnt == 1 indicates that the mixnum spanning tree is builded sucessfully.
		for (int i = 0; i < i_e && cnt != 1; ++i)
		{
			if (merge(edges[i].x, edges[i].y))res += edges[i].w;
		}

		printf("%.2lf\n", res);
	}
	return 0;
}

prim

#include <cstdio>
#include <cmath>
#include <string>
#include <climits>

using namespace std;

const int N = 105;

struct Point
{
	double x;
	double y;
};

Point point[N];
double dist[N], graph[N][N];

bool visit[N];
int i_p, i_e;
double res;


double prim(int n)
{
	for (int i = 0; i < n; ++i)
	{
		dist[i] = graph[1][i];
	}

	visit[1] = true;
	res = 0;

	for (int i = 1; i < n; ++i)
	{
		int min_label;
		double minx = INT_MAX * 1.0;
		for (int j = 0; j < n; ++j)
		{
			if (visit[j])continue;
			if (minx > dist[j])
			{
				minx = dist[j];
				min_label = j;
			}
		}

		visit[min_label] = true;
		res += minx;

		for (int j = 0; j < n; ++j)
		{
			if (!visit[j])
			{
				if (dist[j] > graph[min_label][j])
				{
					dist[j] = graph[min_label][j];
				}
			}
		}
	}
	return res;
}

void init(int n)
{
	memset(visit, 0, sizeof(visit));
}

int main()
{
	int n;
	double dx, dy;

	while (scanf("%d", &n) != EOF)
	{
		i_e = i_p = 0;
		res = 0;

		memset(visit, 0, sizeof(visit));

		for (int i = 0; i < n; ++i)
		{
			scanf("%lf %lf", &dx, &dy);
			point[i_p].x = dx;
			point[i_p].y = dy;
			++i_p;
		}

		for (int i = 0; i < n; ++i)
		{
			for (int j = i + 1; j < n; ++j)
			{
				
				double dd = (point[i].x - point[j].x) * (point[i].x - point[j].x);
				dd += (point[i].y - point[j].y) * (point[i].y - point[j].y);
				graph[i][j] = graph[j][i] = sqrt(dd);
			}
		}

		res = prim(n);

		printf("%.2lf\n", res);
	}
	return 0;
}

HDU 1162 Eddy's picture（图论-最小生成树）

Eddy's picture

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