题目大意:多组数据,每组给一个n(1=<n<=8000),下面有n行,每行有l,r,color(1=<color<=8000),表示将l~r颜色变为color,最后求各种颜色(1~8000)所占区段数,如果所占区段为0则不用输出. 解题思路:基本还是跟区间染色问题差不多,但要注意一些东西 ①这里l,r指的是端点.比如 3 1 4 1 1 2 2 3 4 3 这组数据最后输出结果为: 1 1 2 1 3 1 原因是还有[2,3]这一段颜色为1,处理办法是将更新函数写成(1,l+1,r…

任意门:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1610 Count the Colors Time Limit: 2 Seconds      Memory Limit: 65536 KB Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent…

Count the Colors Time Limit: 2 Seconds      Memory Limit: 65536 KB Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones. Your task is counting the segments of different colors you can s…

Count the Colors Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 1610 Description Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent o…

题目链接 题意 : 一根木棍,长8000,然后分别在不同的区间涂上不同的颜色,问你最后能够看到多少颜色,然后每个颜色有多少段,颜色大小从头到尾输出. 思路 :线段树区间更新一下,然后标记一下,最后从头输出. //ZOJ 1610 #include <cstdio> #include <cstring> #include <iostream> using namespace std ; *],lz[*] ,hashh[*],hash1[*]; //void pushup(…

Count the Colors Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1610 Description Painting some colored segments on a line, some previously painted segments may be covered by some the subseque…

Count the Colors Time Limit: 2 Seconds      Memory Limit: 65536 KB Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones. Your task is counting the segments of different colors you can s…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=610  Count the Colors Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 1610 Description Painting some colored segments on a line, some pre…

<题目链接> 题目大意: 在[0,8000]这个区间内,不断进行一些操作,将其中的一些区间染成特定颜色,如果区间重复的话,后面染的色块会覆盖前面染的色块,问最终[0,8000]这个区间内每种颜色的色块数量是多少. 解题分析: 首先要注意,这是对区间进行更新,.所以update的时候是对输入区间[a,b]的左闭右开或者是左开右闭进行处理.然后本题思路非常清晰,就是按照输入顺序更新线段树对应区间,然后对[0,8000]这个区间从左向右进行色块的统计.PS:虽然最后还是要将所有lazy下放到所有的叶…

题意 : 给出 n 个染色操作,问你到最后区间上能看见的各个颜色所拥有的区间块有多少个 分析 : 使用线段树成段更新然后再暴力查询总区间的颜色信息即可,这里需要注意的是给区间染色,而不是给点染色,所以对于区间(L, R)我们只要让左端点+1即可按照正常的线段树操作来做. #include<bits/stdc++.h> #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 using namespace std; +…