hdu 2461 Rectangles

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HDU 2461 Rectangles#容斥原理

http://acm.hdu.edu.cn/showproblem.php?pid=2461 题目很简单,但是由于询问数M可以很大,所以容易超时,这道题学到了在结构体里面写函数的方法,这样子效率更高,否则的话,这道题就TLE了. 根据容斥原理,先把每个小长方形的面积加上,然后看有没有与该小长方形相交的,用dfs实现,当相交面积为0时,则不进行dfs,且同样遵循奇加偶减(但代码里因为是以第二个作为depth=1开始进行dfs的,所以是奇减偶加). AC代码: #include<iostream>…

求矩形的并矩形个数 1...20m次询问回答要求的r个矩形的并容斥原理dfs优化: 遇到面积交为0时这个dfs分支可以不下去了 #include <iostream> #include <string> #include<sstream> #include <cmath> #include <map> #include <stdio.h> #include <string.h> #include <algori…

hdu 2461(AC) & poj 3695(TLE)(离散化+矩形并)

Rectangles Time Limit: 5000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1503 Accepted Submission(s): 777 Problem Description You are developing a software for painting rectangles on the screen. The softwa…

HDU 2056 Rectangles

Rectangles Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 19110 Accepted Submission(s): 6187 Problem Description Given two rectangles and the coordinates of two points on the diagonals of ea…

HDU 2461 线段树扫描线

给出N个矩形,M次询问每次询问给出R个.问这R个矩形围成的面积经典扫面线求面积并,对每次询问的R个点离散化一下 #include "stdio.h" #include "string.h" #include "algorithm" #include "map" using namespace std; map<int,int>mp; struct P { int x1,y1,x2,y2; }p[21]; str…

HDU 1892 See you~ (二维树状数组)

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1892 See you~ Problem Description Now I am leaving hust acm. In the past two and half years, I learned so many knowledge about Algorithm and Programming, and I met so many good friends. I want to say so…