POJ-2031(Building a Space Station) Kruskal

最新推荐文章于 2019-09-25 09:15:17 发布

YancyKahn

最新推荐文章于 2019-09-25 09:15:17 发布

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分类专栏：图论生成树文章标签： kruskal

本文链接：https://blog.youkuaiyun.com/qq_37753409/article/details/79519644

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本文详细介绍了如何使用最小生成树算法解决POJ-2031问题，通过Kruskal算法实现节点间边的选取，并采用三维空间中球体距离计算方法来确定边的权重。

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POJ-2031

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

struct _Node{
    double x, y, z, r;
    _Node(){}
    _Node(double _x, double _y, double _z, double _r): x(_x), y(_y), z(_z), r(_r){};
};

struct _Edge{
    int from, to;
    double dist;
    _Edge(){};
    _Edge(int _from, int _to, double _dist): from(_from), to(_to), dist(_dist){};
};
const int MAXN = 105;
const int MAXM = 105*104;
int f[MAXN];
_Edge edge[MAXM];
_Node node[MAXN];

int find(int x){
    return f[x] == -1 ? x : f[x] = find(f[x]); 
}

bool cmp(_Edge e1, _Edge e2){
    return e1.dist < e2.dist;
}
double Kruskal(int cnt, int n){
    memset(f, -1, sizeof(f));
    sort(edge, edge + cnt, cmp);
    double res = 0;
    int t = 0;
    for(int i = 0; i < cnt; ++i){
        int a = find(edge[i].from);
        int b = find(edge[i].to);
        if(a != b){ 
            f[a] = b;
            res += edge[i].dist;
            t++;
        }
        if(t == n - 1) break;
    }
    if(t < n - 1) return -1;
    else return res;
}

double getDist(int i, int j){
    double dx = node[i].x - node[j].x;
    double dy = node[i].y - node[j].y;
    double dz = node[i].z - node[j].z;
    double dist = sqrt(dx*dx + dy*dy + dz*dz);
    double t = dist - node[i].r - node[j].r;
    if(t <= 0) return 0;
    else return t;
}
int main(){
    int n;
    while(cin >> n && n){
        for(int i = 0; i < n; ++i) cin >> node[i].x >> node[i].y >> node[i].z >> node[i].r;
        int cnt = 0;
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j) {
                if(i != j){
                    edge[cnt].from = i;
                    edge[cnt].to = j;
                    edge[cnt++].dist = getDist(i, j);
                }
            }
        printf("%.3f\n", Kruskal(cnt, n));
    }
    return 0;
}