跟前面某个题一样,都是区间染色问题,还是用我的老方法,区间离散化+二分区间端点+区间处理做的,时间跑的还挺短 坑爹的情况就是最左端是0,最右端是1e9,区间求的是开区间 #include <stdio.h> #include <string.h> #include <math.h> #include <algorithm> using namespace std; typedef struct { int l; int r; bool color; }seg…

1019 离散化都忘记怎么写了 注意两个端点 离散化后用线段树更新区间 混色为-1  黑为2  白为1  因为N不大 最后直接循环标记这一段的颜色查找 #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> using namespace std; #define N 100010 #define LL long l…

Line Painting Time limit: 2.0 secondMemory limit: 64 MB The segment of numerical axis from 0 to 109 is painted into white color. After that some parts of this segment are painted into black, then some into white again and so on. In total there have b…

2025. Line Fighting Time limit: 1.0 second Memory limit: 64 MB Boxing, karate, sambo- The audience is sick of classic combat sports. That is why a popular sports channel launches a new competition format based on the traditional Russian entertainment…

题目大意;说是可以吧一段区间变成白色或者黑色, 区间(0-10^9)初始都是白色,问经过n次操作以后最大的连续白色区间 Problem Description The segment of numerical axis from 0 to 109 is painted into white color. After that some parts of this segment are painted into black, then some into white again and so o…

原文地址:http://blog.csdn.net/zearot/article/details/48299459(如有侵权,请联系博主,立即删除.) 线段树详解    By 岩之痕 目录: 一:综述     二:原理    三:递归实现    四:非递归原理      五:非递归实现 六:线段树解题模型    七:扫描线   八:可持久化 (主席树)     九:练习题 一:综述 假设有编号从1到n的n个点,每个点都存了一些信息,用[L,R]表示下标从L到R的这些点. 线段树的用处就是,对编号…

前言:前一段时间在网上找了一个线段树题目列表,我顺着做了一些,今天我把做过的整理一下.感觉自己对线段树了解的还不是很深,自己的算法能力还要加强.光练代码能力还是不够的,要多思考.向队友学习,向大牛学习. ZOJ1610 题目大意:先后对线段涂色,最后统计每种颜色出现的段数,为0则不输出. 分析:以点建树,每个节点有一个标记col,初始为-1,表示未涂过色.涂色时,若完全覆盖,则col为颜色编号,若覆盖一部分,则先将标记下放到左右儿子节点,再将标号标记为-2,表示此节点被部分覆盖过.最后从根节点开…

http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DSL_5_A The Maximum Number of Customers Idea: find the most overlapped segment.  Similiar to the line painting Mark the left and right element, x: L, O: R #include<cstdio> #include<cstdlib&…

http://acm.hust.edu.cn/vjudge/contest/126149#problem/H 给定一条二次函数 f (x) = a * x * x + b * x + c 求一个最小的k,使得f(x) + f(x + 1) + f(x + 2) ..... + f(x + k - 1) 不等于 0 恒成立. 首先把参数统一写成 x + t这样的形式,然后带入去 化简有:a*x*x + (2*a*t+b)*x + a*t*t+b*t+c //现在的t是从0--k-1 列出k个式子,…

Codeforces Round #256 (Div. 2) C C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output Bizon the Champion isn't just attentive, he also is very hardworking. Bizon the Champion d…