http://acm.hdu.edu.cn/showproblem.php?pid=5023 Problem Description Corrupt governors always find ways to get dirty money. Paint something, then sell the worthless painting at a high price to someone who wants to bribe him/her on an auction, this seem…

A Corrupt Mayor's Performance Art Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others) Problem Description Corrupt governors always find ways to get dirty money. Paint something, then sell the worthless painting at a…

A Corrupt Mayor's Performance Art Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)Total Submission(s): 33    Accepted Submission(s): 11 Problem Description Corrupt governors always find ways to get dirty money. Pa…

Link:  http://acm.hdu.edu.cn/showproblem.php?pid=5023 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <vector> #include <string> #include <cmath> using namespace std; typedef…

题意: n,q<=1e5,a[i],b[i][j]<=1e9,保证能力值互不相同,询问之间保留前面的影响 思路:其实把大于a[1]的看成0,小于的看成1,设第i天小于a[1]的有b[i]个,本质上就是这样一个过程: 刚开始有b[0]个小于a[1]的,第1天先减去r[1]看是否小于0,若小于0则结束,再加上b[1],以此类推 由此可见每次询问只是单点修改了b[x],判断的话就是在判最小的前缀和是否<0 用线段树维护一下最小的前缀和即可 #include<bits/stdc++.h&g…

A Corrupt Mayor's Performance Art Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others) Total Submission(s): 1905    Accepted Submission(s): 668 Problem Description Corrupt governors always find ways to get dirty money…

Problem Description Corrupt governors always find ways to get dirty money. Paint something, then sell the worthless painting at a high price to someone who wants to bribe him/her on an auction, this seemed a safe way for mayor X to make money. Becaus…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5023 解题报告:一面墙长度为n,有N个单元,每个单元编号从1到n,墙的初始的颜色是2,一共有30种颜色,有两种操作: P a b c  把区间a到b涂成c颜色 Q a b 查询区间a到b的颜色 线段树区间更新,每个节点保存的信息有,存储颜色的c,30种颜色可以压缩到一个int型里面存储,然后还有一个tot,表示这个区间一共有多少种颜色. 对于P操作,依次往下寻找,找要更新的区间,找到要更新的区间之前…

题意:给定一个1-n的墙,然后有两种操作,一种是P l ,r, a 把l-r的墙都染成a这种颜色,另一种是 Q l, r 表示,输出 l-r 区间内的颜色. 析:应该是一个线段树+状态压缩,但是我用set暴力过去了.用线段树+状态压缩,区间更新,很简单,就不说了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #i…

题目原文废话太多太多太多,我就不copyandpaste到这里啦..发个链接吧题目 题目意思就是:P  l  r  c  将区间 [l ,r]上的颜色变成c    Q  l r 就是打印出区间[l,r]中所有的颜色,并且要升序排列出来,注意最坑的就是各个区间的颜色初始化为2,这个条件竟然夹杂在那又长又臭的题目的某个角落里面!! 比赛的时候思路是有的,并且也能想到用set来撸,哎,对set的用法太挫逼了,写线段树又写得太挫逼了,后来补回这道题的时候,才发现其实是一道非常常规的线段树,所以最近给自己…