D. The Union of k-Segments time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output You are given n segments on the coordinate axis Ox and the number k. The point is satisfied if it belongs to at l…

题意: 给你n个线段和一个整数k,你需要找出来所有能被任意k条线段同时覆盖的区间个数的最小值,并按从左到右的顺序输出每个区间. 题解: 对于题目输入的n个线段的左端点L,右端点R,把它们分开放在结构体里面.结构体的值为右端点或者左端点的值,如果这个点是右端点就标记为0,否则标记为1. 然后对结构体排序,如果结构体的值相等就把标记为0的结构体放在前面,否则就按照值从小到大排序 然后从0开始对结构体进行遍历,如果遇到右端点那么线段重合数量cnt加1,如果遇到左端点,那么肯定会有一个线段到达了使用的最…

最先想到的是把两个linked lists 合并成一个. 这样从第一个开始一个一个吞并,直到所有list都被合并. class ListNode:# Definition for singly-linked list. def __init__(self, x): self.val = x self.next = None class Solution: # @param {ListNode[]} lists # @return {ListNode} def mergeKLists_0(self…

本文出处:http://www.cnblogs.com/wy123/p/7884986.html 周围又有人在讨论UNION和UNION ALL,对于UNION和UNION ALL,网上说的最多的就是性能问题(实在不想说出来这句话:UNION ALL比UNION快)其实根本不想炒UNION和UNION ALL这碗剩饭了,每次看到网上说用这个不用那个,列举的一条一条的那种文章,只要看到说UNION ALL比UNION性能好的就…… 对于合并的结果集,UNION是去重的,UNION ALL是不去重的…

Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is rebellious and obeys his parents by simply throwing his toys in…

D. Vika and Segments 题目连接: http://www.codeforces.com/contest/610/problem/D Description Vika has an infinite sheet of squared paper. Initially all squares are white. She introduced a two-dimensional coordinate system on this sheet and drew n black hor…

D. Vika and Segments     Vika has an infinite sheet of squared paper. Initially all squares are white. She introduced a two-dimensional coordinate system on this sheet and drew n black horizontal and vertical segments parallel to the coordinate axes.…

题目链接:http://codeforces.com/contest/610/problem/D 就是给你宽度为1的n个线段,然你求总共有多少单位的长度. 相当于用线段树求面积并,只不过宽为1,注意y和x的最大都要+1,这样才相当于求面积. #include <iostream> #include <cstdio> #include <cstring> #include <map> #include <algorithm> using names…

可以转变成上一题(hdu1542)的形式,把每条线段变成宽为1的矩形,求矩形面积并 要注意的就是转化为右下角的点需要x+1,y-1,画一条线就能看出来了 #include<bits/stdc++.h> #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #pragma comment(linker,…

还记得lyf说过k=2的方法,但是推广到k是其他的话有点麻烦.现在这里采取另外一种方法. 先将所有线段按照L进行排序,然后优先队列保存R的值,然后每次用最小的R值,和当前的L来维护答案即可.同时,如果Q的size()比k大,那么就弹出最小的R. 具体见代码: #include <stdio.h> #include <algorithm> #include <string.h> #include <set> #include <vector> #i…