1012: A MST Problem

1012: A MST Problem

时间限制: 1 Sec  内存限制: 32 MB
提交: 63  解决: 33
[提交][状态][讨论版][命题人:外部导入]

题目描述

It is just a mining spanning tree ( 最小生成树 ) problem, what makes you a little difficult is that you are in a 3D space.

输入

The first line of the input contains the number of test cases in the file. And t he first line of each case

contains one integer numbers n(0<n<30) specifying the number of the point . The n next n line s, each line

contain s Three Integer Numbers xi,yi and zi, indicating the position of point i.

输出

For each test case, output a line with the answer, which should accurately rounded to two decimals .

样例输入

2
2
1 1 0
2 2 0
3
1 2 3
0 0 0
1 1 1

样例输出

1.41
3.97

#include<iostream>
#include<cmath>
#include<cstdio>
#define INF 99999999
#define MAXV 100
using namespace std;
typedef struct
{
	double edges[100][100];
	int n;
	int e;
}MatGraph;
struct point
{
	int x;
	int y;
	int z;
}point[100];

MatGraph g;

void CreateMat(MatGraph &g, double graph[100][100], int n)
{
	int i, j;
	g.n = n;
	for(i = 0; i < g.n; ++i)
	{
		for(j = 0; j < g.n; ++j)
		{
			g.edges[i][j] = graph[i][j];
		}
	}
}

double Prim(MatGraph g, int v)
{
	double lowcost[100];
	double min_ = INF;
	double r = 0;
	int i, j, k;
	int n = g.n;
	for(i = 0; i < n; ++i)
	{
		lowcost[i] = g.edges[v][i];
	}
	for(i = 1; i < n; ++i)
	{
		min_ = INF;
		for(j = 0; j < n; ++j)
		{
			if(lowcost[j] != -1 && lowcost[j] < min_)
			{
				min_ = lowcost[j];
				k = j;
			}
		}
		lowcost[k] = -1;
		r = r + min_;
		for(j = 0; j < n; ++j)
		{
			if(g.edges[k][j] < lowcost[j] && g.edges[k][j] != -1)
			{
				lowcost[j] = g.edges[k][j];
			}
		}
	}
	return r;
}

int main()
{
	double length, ans, result;
	double graph[100][100];
	int num, i, j, k, l, t, n;
	cin >> t;
	while(t--)
	{
		cin >> n;
		for(i = 0; i < n; ++i)
		{
			for(j = 0; j < n; ++j)
			{
				graph[i][j] = INF;
			}
		}
		for(i = 0; i < n; ++i)
		{
			cin >> point[i].x >> point[i].y >> point[i].z;
		}
		for(i = 0; i < n; ++i)
		{
			for(j = 0; j < n; ++j)
			{
				length = sqrt((point[i].x - point[j].x)*(point[i].x - point[j].x) + (point[i].y - point[j].y)*(point[i].y - point[j].y) + (point[i].z - point[j].z)*(point[i].z - point[j].z));
				graph[i][j] = graph[j][i] = length;
			}
		}
		for(i = 0; i < n; ++i)
		{
			for(j = 0; j < n; ++j)
			{
				if(i == j)
				{
					graph[i][j] = -1;
				}
			}
		}
		CreateMat(g, graph, n);
		result = Prim(g, 0);
		printf("%.2lf
", result);
	}
	return 0;
}