zoukankan      html  css  js  c++  java
  • poj2420 A Star not a Tree?

    传送门:http://poj.org/problem?id=2420

    【题解】

    费马点问题,模拟退火。%.0f神坑

    # include <math.h>
    # include <stdio.h>
    # include <stdlib.h>
    # include <string.h>
    # include <iostream>
    # include <algorithm>
    // # include <bits/stdc++.h>
    
    using namespace std;
    
    typedef long long ll;
    typedef long double ld;
    typedef unsigned long long ull;
    const int M = 5e5 + 10;
    const int mod = 1e9+7;
    const double pi = acos(-1.0);
    
    # define RG register
    # define ST static
    
    double minx, maxx, miny, maxy;
    int n;
    struct pa {
        double x, y;
        double dis;
        pa() {}
        pa(double x, double y, double dis) : x(x), y(y), dis(dis) {}
    }a[M];
    
    namespace SA {
        const double eps = 1e-2, DEC = 0.9, ACCEPT_DEC = 0.5;
        const int N = 30, T = 30, RAD = 1000; 
        inline double rand01() {
            return rand() % (RAD + 1) / (1.0 * RAD);
        }
        inline double getdist(double x, double y) {
            double ret = 0;
            for (int i=1; i<=n; ++i)
                ret += sqrt((x-a[i].x)*(x-a[i].x)+(y-a[i].y)*(y-a[i].y));
            return ret;
        }
        inline pa randpoint(double px, double py, double qx, double qy) {
            double x = (qx - px) * rand01() + px, y = (qy - py) * rand01() + py;
            return pa(x, y, getdist(x, y)); 
        }
        pa res[N + 5];
        inline double main() {
            res[1] = pa((minx+maxx)/2, (miny+maxy)/2, getdist((minx+maxx)/2, (miny+maxy)/2));
            for (int i=2; i<=N; ++i) {
                double x = rand01() * (maxx - minx) + minx;
                double y = rand01() * (maxy - miny) + miny;
                res[i] = pa(x, y, getdist(x, y));
            }
            double temper = (maxx+maxy-minx-miny)*n, accept = 0.6;
            while(temper > eps) {
                for (int i=1; i<=N; ++i)
                    for (int j=1; j<=T; ++j) {
                        pa t = randpoint(max(res[i].x - temper, minx), max(res[i].y - temper, miny), min(res[i].x + temper, maxx), min(res[i].y + temper, maxy));
                        if(t.dis < res[i].dis) res[i] = t;
                        else if(rand01() <= accept) res[i] = t;
                    }
                temper *= DEC;
                accept *= ACCEPT_DEC;
            }
            double dis = 1e18;
            for (int i=1; i<=N; ++i) 
                if(res[i].dis < dis) dis = res[i].dis;
            return dis;
        }
    }
    
    int main() {
        srand(19260817); 
        while(cin >> n) {
            maxx = maxy = -1e18, minx = miny = 1e18;
            for (int i=1; i<=n; ++i) {
                scanf("%lf%lf", &a[i].x, &a[i].y);
                maxx = max(maxx, a[i].x);
                minx = min(minx, a[i].x);
                maxy = max(maxy, a[i].y);
                miny = min(miny, a[i].y);
            }
            double ans = SA::main();
            printf("%.0f
    ", ans);
        }
    
        return 0;
    }
    View Code
  • 相关阅读:
    HTTP 常用 Header
    三大框架面试笔记-Hibernate
    三大框架面试笔记-Spring
    Bloom Filter概念和原理
    笔记:编写高质量代码 改善Java程序的151个建议
    笔记:Effective Java
    HTTP状态码
    JAVA8 十大新特性详解
    java7 新特性
    笔记:Hadoop权威指南 第10章 管理Hadoop
  • 原文地址:https://www.cnblogs.com/galaxies/p/poj2420.html
Copyright © 2011-2022 走看看