YTU 1012: A MST Problem

1012: A MST Problem

时间限制: 1 Sec  内存限制: 32 MB

提交: 7  解决: 4

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

你  离  开  了  ，  我  的  世  界  里  只  剩  下  雨  。  。  。

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;
const int infinity=99999999;
const int maxnum=105;
double map1[maxnum][maxnum];
bool visited[maxnum];
double low[maxnum];
int nodenum;
struct node
{
    int x;
    int y;
    int z;
} nd[105];
double prim()
{
    int i,j,pos=1;
    double result,Min;
    memset(visited,0,sizeof(visited));//初始化都未标记
    result=0;
    for(i=1; i<=nodenum; i++)
        low[i]=map1[pos][i];
    visited[pos]=1;//把1号作为起点
    for(i=2; i<=nodenum; i++) //这个i没有其他的意思就是一个循环次数
    {
        Min=infinity;
        pos=-1;//从1号开始找最小的边
        for(j=1; j<=nodenum; j++)
            if(!visited[j]&&Min>low[j])
            {
                Min=low[j];
                pos=j;
            }
        if(pos==-1)
            return -1;
        visited[pos]=1;//做到与1os号最近的边
        result+=Min;//加权值
        for(j=1; j<=nodenum; j++)
            if(!visited[j]&&low[j]>map1[pos][j])
                low[j]=map1[pos][j];//这个就是替换未被标记的最小权值！
    }
    return result;
}
int main()
{
    int n,i,j,t;
    double lenth,ans;
    cin>>t;
    while(t--)
    {
        cin>>n;
        nodenum=n;
        for(i=1; i<=nodenum; i++)
            for(j=1; j<=nodenum; j++)
                map1[i][j]=infinity;
        for(i=1; i<=n; i++)
            cin>>nd[i].x>>nd[i].y>>nd[i].z;
        for(i=1; i<=n; i++)
        {
            for(j=i+1; j<=n; j++)
            {
                lenth=sqrt((nd[i].x-nd[j].x)*(nd[i].x-nd[j].x)+(nd[i].y-nd[j].y)*(nd[i].y-nd[j].y)+(nd[i].z-nd[j].z)*(nd[i].z-nd[j].z));
                //  printf("djklsajiofgioj%.2lf\n",lenth);
                map1[i][j]=map1[j][i]=lenth;
            }
        }
        ans=prim();//开始rim算法
        if(ans==-1)
            cout<<"?"<<endl;
        else
            printf("%.2lf\n",ans);
    }
}