YTU 1012: A MST Problem

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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

221 1 02 2 031 2 30 0 01 1 1

样例输出

1.413.97

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);    }}

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