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…

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) Total Submission(s): 1905    Accepted Submission(s): 668 Problem Description Corrupt governors always find ways to get dirty money…

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…

题意:有一个长板子,分成多段,有两种操作,第一种是C给从a到b那段染一种颜色c,另一种是P询问a到b有多少种不同的颜色.Sample Input2 2 4  板长 颜色数目 询问数目C 1 1 2P 1 2C 2 2 2P 1 2Sample Output21 sum用二进制记录区间内颜色状态,col记录染上的颜色(其他颜色会被染上的颜色完全覆盖) 2015-07-23:专题复习 #include<cstdio> #include<iostream> #include<alg…

题目分析:线段树区间更新+离散化 代码如下: # include<iostream> # include<cstdio> # include<queue> # include<vector> # include<list> # include<map> # include<set> # include<cstdlib> # include<string> # include<cstring&g…

http://poj.org/problem?id=2528 https://www.luogu.org/problem/UVA10587 Description The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim…

题目链接: 传送门 Mayor's posters Time Limit: 1000MS     Memory Limit: 65536K Description The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim…

题意:给一个区间,表示这个区间贴了一张海报,后贴的会覆盖前面的,问最后能看到几张海报. 思路: 之前就不会离散化,先讲一下离散化:这里离散化的原理是:先把每个端点值都放到一个数组中并除重+排序,我们就得到了处理后的数组,现在我们只需要用二分查找端点值在整个数组的下标,这样就达到了离散化的目的,压缩了长度.因为这里很特殊,不能用一般的离散化去做,如果区间两端只差1,那么需要给这个区间再加一个值,这个其他题解讲到了. 这里的区间更新和上一题不太一样,有一些地方要注意一下 代码: #include<q…