[POJ1328]Radar Installation 试题描述 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 onl…

Radar Installation 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…

题目:Radar Installation 对于x轴上方的每个建筑 可以计算出x轴上一段区间可以包含这个点 所以就转化成 有多少个区间可以涵盖这所有的点 排序之后贪心一下就ok 用cin 好像一直t看了好多blog 改了scanf 过了 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> #include<cmat…

Radar Installation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 54593   Accepted: 12292 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 loca…

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…

https://vjudge.net/problem/POJ-1328 贪心策略选错了恐怕就完了吧.. 一开始单纯地把island排序,然后想从左到右不断更新,其实这是错的...因为空中是个圆弧. 后来知道了通过对每个island求雷达范围,将这些范围排序,但是去重策略也没想对.因为有两种情况,都需要更新t值,但是只有一种需要ans++. #include<iostream> #include<cstdio> #include<queue> #include<cs…

题目链接. 题意: 给定一坐标系,要求将所有 x轴 上面的所有点,用圆心在 x轴, 半径为 d 的圆盖住.求最少使用圆的数量. 分析: 贪心. 首先把所有点 x 坐标排序, 对于每一个点,求出能够满足的 最靠右的圆心,即雷达的位置. 要保证雷达左面的点都被覆盖,如果不能覆盖就向左移,移到能将左边未覆盖的覆盖.如果后面的店不在雷达的覆盖区,则再加一雷达. #include <iostream> #include <cstdio> #include <cstdlib> #i…

对每个岛屿,能覆盖它的雷达位于线段[x-sqrt(d*d-y*y),x+sqrt(d*d+y*y)],那么把每个岛屿对应的线段求出来后,其实就转化成了经典的贪心法案例:区间选点问题.数轴上有n个闭区间[ai,bi],取尽量少的点,使得每个区间内都至少有一个点.选法是:把区间按右端点从小到大排序(右端点相同时按左端点从大到小),然后每个没选的区间选最右边的一个点即可. #include<iostream> #include<cstdio> #include<cstdlib>…

题目链接:http://poj.org/problem?id=1328 题解:区间选点类的题目,求用最少的点以使得每个范围都有点存在.以每个点为圆心,r0为半径,作圆.在x轴上的弦即为雷达可放置的范围.这个范围可用勾股定理求得.记录每个点的范围,然后排序,贪心. (由于不熟悉qsort的用法,折腾了一个小时.后来清晰了:qsort 的cmp是通过正负判断(但结构体又不是),而sort的cmp是通过对错判断,即0或1.所以qsort的cmp用‘-’(结构体除外),sort的cmp用‘<’和‘>’…

ZOJ地址:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=360 POJ地址:http://poj.org/problem?id=1328 解题思路:贪心 ];                  : -;  }         ;                     && r == ){                       }          lose = ;          d = (        …