翻出一年多前的代码看,发现以前的代码风格很糟糕 题意:给你n个点 m为圆的半径,问需要多少个圆能把全部点圈到 #include <iostream> #include <algorithm> #include <cmath> using namespace std; struct ss{ double x,y; }a[1005]; int cmp(ss x,ss y) { if(x.x==y.x)return x.y<y.y; else return x.x<…

POJ 1328 Radar Installation https://vjudge.net/problem/POJ-1328 题目: Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation…

http://poj.org/problem?id=1328 思路: 1.肯定y大于d的情况下答案为-1,其他时候必定有非负整数解 2.x,y同时考虑是较为麻烦的,想办法消掉y,用d^2-y^2获得圆心允许范围,问题转化为在许多圆心允许范围内取尽可能少的点,也即在许多线段上取尽可能少的点,使得所有线段上都有点被取到 3.从左往右考虑,完全在左边的线段肯定要取点,如果这个点在当前线段上已经取了,明显就可以忽略当前线段,明显在线段上的最优点是右端点 #include <iostream> #inc…

题目:http://poj.org/problem?id=1328   题意:建立一个平面坐标,x轴上方是海洋,x轴下方是陆地.在海上有n个小岛,每个小岛看做一个点.然后在x轴上有雷达,雷达能覆盖的范围为d,问至少需要多少个雷达能监测到多有的小岛. 思路:从左到右把每个小岛的放置雷达的区间求出,按结束点排序,从左至右看,当发现下一个区间的起始点大于前面所有区间的最小结束点的时候,答案加一. #include <stdio.h> #include<string.h> #include…

Description Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, s…

Input The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n li…

Description Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, s…

本题是贪心法题解.只是须要自己观察出规律.这就不easy了,非常easy出错. 一般网上做法是找区间的方法. 这里给出一个独特的方法: 1 依照x轴大小排序 2 从最左边的点循环.首先找到最小x轴的圆 3 以这个圆推断能够包含右边的多少个圆,直到不能够包含下一个点,那么继续第2步,画一个新圆. 看代码吧,应该非常清晰直观的了. 效率是O(n),尽管有嵌套循环.可是下标没有反复.一遍循环就能够了.故此是O(n). #include <stdio.h> #include <cmath>…

(-￣▽￣)-* #include<iostream> #include<cstdio> #include<algorithm> #include<cmath> using namespace std; struct node { double l,r; //找到以岛为圆心,以d为半径的圆与坐标x轴的左交点l.右交点r //雷达只有设在l~r之间,岛才在雷达覆盖范围内 }a[]; int cmp(node a,node b) { return a.l<…

题目地址:http://poj.org/problem?id=1328 /* 贪心 (转载)题意:有一条海岸线,在海岸线上方是大海,海中有一些岛屿, 这些岛的位置已知,海岸线上有雷达,雷达的覆盖半径知道, 问最少需要多少个雷达覆盖所有的岛屿. (错误)思路:我开始是想从最左边的点雷达能探测的到的最右的位置出发,判断右边其余的点是否与该点距离小于d 是,岛屿数-1:不是,雷达数+1,继续... (正确)思路:每个岛屿的座标已知,以雷达半径为半径画圆,与x轴有两个交点. 也就是说,若要覆盖该岛,雷达…