poj 2031 Building a Space Station (最小生成树)

2016年7月14日 0 作者 CrazyKK

poj 2031

 

题意：三维空间中n个球要相连。。。通路的代价是距离。。。如果球相交（切）或者包含那么不用建通路就能联系。。。问联系所有球的最小代价。。。

思路：裸的最小生成树。。。。先预处理球和球表面的距离。。。距离是负数的处理成0.。。然后mst搞之。。。不算CE的话是1A….

 

 

C++
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/* ***********************************************
Author :111qqz
Created Time :2016年07月14日 星期四 15时44分53秒
File Name :code/poj/2031.cpp
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const  int N =105;
int n;
int m;
double a[N][N];
int f[N];
int dblcmp(double d)
{
    return d<-eps?-1:d>eps;
}
void init()
{
    for ( int i = 1 ; i  <= n ; i++) f[i] = i;
}
struct circle
{
    double x,y,z,r;
    double dis(circle b)
    {
        double res = 0 ;
        res = (x-b.x)*(x-b.x)+(y-b.y)*(y-b.y)+(z-b.z)*(z-b.z);
        res = sqrt(res);
        res = res - r - b.r;
        if (dblcmp(res)<0) res = 0 ;
        return res;
    }
    void input()
    {
        scanf("%lf%lf%lf%lf",&x,&y,&z,&r);
    }
}cir[N];
struct Edge
{
    int u,v;
    double w;
    bool operator < (Edge b)const
    {
        return w<b.w;
    }
}edge[N*N/2];
int root ( int x)
{
    if (x!=f[x]) f[x] = root (f[x]);
    return f[x];
}
void merge( int x,int y)
{
    int rx = root (x);
    int ry = root(y);
    if (rx==ry) return;
    f[rx] = ry;
}
int main()
{
#ifndef  ONLINE_JUDGE
    freopen("code/in.txt","r",stdin);
#endif
    while (~scanf("%d",&n))
    {
        if (n==0) break;
        init();
        for ( int i = 1 ;i  <= n ; i++) cir[i].input();
        for ( int i = 1 ; i <= n ; i++)
            for ( int j = 1 ; j <= n ; j++)
                a[i][j]=a[j][i]=cir[i].dis(cir[j]);
        m = 0 ;
        for ( int i = 1 ; i <= n ; i++)
            for ( int j = 1 ; j <= n ; j++)
                if (i<j)
                {
                    m++;
                    edge[m].u = i ;
                    edge[m].v = j;
                    edge[m].w = a[i][j];
                }
        sort(edge+1,edge+m+1);
        double mst = 0 ;
        int cnt = 0 ;
        for ( int i = 1 ; i <= m ; i++)
        {
            int u = edge[i].u;
            int v = edge[i].v;
            double w = edge[i].w;
            if (root(u)==root(v)) continue;
            merge(u,v);
            mst +=w;
            cnt++;
            if (cnt>=n-1) break;
        }
        if (dblcmp(mst)==0)
        {
            puts("0.000");
        }else
        {
            printf("%.3f\n",mst);
        }
    }
#ifndef ONLINE_JUDGE  
    fclose(stdin);
#endif
    return 0;
}
 
 
 
 

分类目录poj
标签mst

关于作者

ex-ACMer@hust，stackoverflow-engineer@sensetime

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