POJ1328Radar Installation】的更多相关文章

Radar Installation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 68597   Accepted: 15373 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…
Radar Installation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 64472   Accepted: 14497 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…
http://poj.org/problem?id=1328 题的大意就是说在海里有小岛,坐标位置会给出,需要岸边的雷达覆盖所有的小岛,但雷达的覆盖范围有限,所以,需要最少的雷达覆盖所有的小岛,但若是有小岛没法被雷达给覆盖到,就输出-1: 这个题的话可以转化成区间问题就是看雷达的覆盖范围作为半径,A若是小岛的位置,根据雷达的覆盖范围只要不小于这个点的Y坐标,那个覆盖范围就是这个三角形的斜边,所以只要雷达位于1,2边上就可以覆盖到这个小岛 #include<cstdio> #include<…
对于每一个点,可以找到他在x轴上的可行区域,这样的话就变为了对区间的贪心. #include<iostream> #include<stdio.h> #include<string.h> #include<map> #include<vector> #include<set> #include<stack> #include<queue> #include<algorithm> #include&l…
[思路] 以每一座岛屿为圆心,雷达范围为半径作圆,记录下与x轴的左右交点.如果与x轴没交点,则直接退出输出“-1”.以左交点为关键字进行排序,从左到右进行贪心.容易知道,离每一个雷达最远的那一座岛与雷达相距恰巧为半径的时候,可以得到最优解.假设上一个雷达与第before座岛相距为半径大小,对于当前的岛屿i: 如果before岛的右交点在i岛左交点的左侧,此时必然需要一个新的雷达,将当前before暂定为i,雷达数加一: 如果before岛的右交点在i岛右交点的右侧,为了让雷达举例尽可能地广,将雷…
原题链接 经典贪心,转化为问题为,对于所有的区间,求最小的点数能使每个区间都至少有一个点. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<queue> using namespace std; const int MAXN=100000+5; const double eps=0.00…
Took me a while to suffer from the first successful souce code installation of mysql-5.6.34. Just put it here and share it with u. Env.OS:Red Hat Enterprise Linux Server release 5.5 (Tikanga)Mysql: mysql-5.6.341. Install Cmake [root@exadata1 cmake-3.…
Create an offline installation of Visual Studio 2017 RC ‎2016‎年‎12‎月‎7‎日                                                                                                  In this article Create a layout Install from a layout Update an installation lay…
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