ZOJ 1610 Count the Colors(线段树,区间覆盖,单点查询)
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 see at last.
Input
The first line of each data set contains exactly one integer n, 1 <= n <=
8000, equal to the number of colored segments.
Each of the following n lines consists of exactly 3 nonnegative integers separated
by single spaces:
x1 x2 c
x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicates
the color of the segment.
All the numbers are in the range [0, 8000], and they are all integers.
Input may contain several data set, process to the end of file.
Output
Each line of the output should contain a color index that can be seen from the
top, following the count of the segments of this color, they should be printed
according to the color index.
If some color can't be seen, you shouldn't print it.
Print a blank line after every dataset.
Sample Input
5
0 4 4
0 3 1
3 4 2
0 2 2
0 2 3
4
0 1 1
3 4 1
1 3 2
1 3 1
6
0 1 0
1 2 1
2 3 1
1 2 0
2 3 0
1 2 1
Sample Output
1 1
2 1
3 1
1 1
0 2
1 1
用col[rt]记录区间的颜色,-1表示多个颜色。。。
#include <bits/stdc++.h>
using namespace std;
#define root 1,n,1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define lr rt<<1
#define rr rt<<1|1
const int N = ;
int n , m , col[N<<],tag[N<<] ; void Down( int l , int r , int rt ) {
if( col[rt] != - ) col[lr] = col[rr] = col[rt] ;
}
void Up( int rt ){
if( col[lr] == col[rr] ) col[rt] = col[lr];
else col[rt] = - ;
}
void build( int l , int r , int rt ){
col[rt] = - ;
if( l == r ) return ;
int mid = (l+r)>>;
build(lson),build(rson);
}
void update( int l , int r , int rt , int L , int R , int c ) {
if( L == l && r == R ) {
col[rt] = c ; return ;
}
if( col[rt] == c ) return ;
else Down( l,r,rt ) , col[rt] = - ;
int mid = (l+r)>>;
if( R <= mid ) update(lson,L,R,c);
else if( L > mid ) update(rson,L,R,c);
else update(lson,L,mid,c) , update(rson,mid+,R,c);
Up(rt);
} int query( int l , int r , int rt , int x ) {
if( col[rt] != - || l == r ) return col[rt];
Down(l,r,rt);
int mid = (l+r)>>;
if( x <= mid ) return query(lson,x);
else return query(rson,x);
} int main()
{
int _ ,x ,y , c ;
n = ;
while( ~scanf("%d",&m) ) {
memset( tag , , sizeof tag );
build(root);
while( m-- ) {
scanf("%d%d%d",&x,&y,&c);x++;
update(root,x,y,c);
}
int last = - , now ;
for( int i = ; i <= n ; ++i ) {
now = query(root,i);
if( last == - ) { last = now ; continue ; }
if( last == now ) continue ;
else tag[last]++; last=now;
}
for( int i = ; i <= n ; ++i ) if( tag[i] ){
printf("%d %d\n",i,tag[i]);
}printf("\n");
}
}
ZOJ 1610 Count the Colors(线段树,区间覆盖,单点查询)的更多相关文章
- zoj 1610 Count the Colors 线段树区间更新/暴力
Count the Colors Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.zju.edu.cn/onlinejudge/show ...
- ZOJ 1610 Count the Color(线段树区间更新)
描述Painting some colored segments on a line, some previously painted segments may be covered by some ...
- ZOJ 1610 Count the Colors (线段树成段更新)
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 ...
- POJ 2528 Mayor's posters(线段树,区间覆盖,单点查询)
Mayor's posters Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 45703 Accepted: 13239 ...
- 【POJ 2777】 Count Color(线段树区间更新与查询)
[POJ 2777] Count Color(线段树区间更新与查询) Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4094 ...
- HDU.1556 Color the ball (线段树 区间更新 单点查询)
HDU.1556 Color the ball (线段树 区间更新 单点查询) 题意分析 注意一下pushdown 和 pushup 模板类的题还真不能自己套啊,手写一遍才行 代码总览 #includ ...
- ZOJ 1610 Count the Colors【题意+线段树区间更新&&单点查询】
任意门:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1610 Count the Colors Time Limit: 2 ...
- ZOJ 1610.Count the Colors-线段树(区间染色、区间更新、单点查询)-有点小坑(染色片段)
ZOJ Problem Set - 1610 Count the Colors Time Limit: 2 Seconds Memory Limit: 65536 KB Painting s ...
- HDU 5861 Road 线段树区间更新单点查询
题目链接: http://acm.split.hdu.edu.cn/showproblem.php?pid=5861 Road Time Limit: 12000/6000 MS (Java/Othe ...
随机推荐
- 结合pychrom与selenium实现页面自动登录
缘起 一直在浏览器里用Katalon插件录制一些常用的流程,以减少重复操作,也就自然而然想自己搞搞自动化测试,但无奈登录一关跨不过去,就无法串起来.(不想让开发添加万能验证码的功能)首先想到的是识别验 ...
- 实验查看PHP本地的Session信息
通过Nginx调度器负载后端两台Web服务器,实现以下目标: - 部署Nginx为前台调度服务器 - 调度算法设置为轮询 - 后端为两台LNMP服务器 - 部署测试页面,查看PHP本地的Session ...
- 发布程序包到Nuget
今天想着别人都把自己做的程序包发布到nuget上去开放给别人使用,那么我是否也能这么干呢,于是就研究了一番,发现还真可以,而且非常简单,接下来就介绍下发布自己的程序包到nuget上的方法. 一.创建公 ...
- idea hibernate console 执行hql报错
报错信息 hql> select a from GDXMZD a[2019-08-29 13:45:01] org.hibernate.service.spi.ServiceException: ...
- Vue 学习之 vue-router2
---恢复内容开始--- 一.路由的安装: npm安装 npm install vue-router --save 执行命令完成vue-router的安装,并在package.json中添加了vue- ...
- 漫谈五种IO模型
阅读目录 1 基础知识回顾 2 I/O模式 3 事件驱动编程模型 网络编程里常听到阻塞IO.非阻塞IO.同步IO.异步IO等概念,搞清楚这些概念之前,还得先回顾一些基础的概念. 1 基础知识回顾 注意 ...
- 父工程 pom版本
<!-- 集中定义依赖版本号 --> <properties> <junit.version>4.12</junit.version> <spri ...
- vue框架搭建--移动端
由于Vue官方提供了vue-cli手脚架,所以快速构建出个简单的项目框架.在做移动端项目时,因为移动端的特性可能会用到些比较常用的插件,就在这里简单介绍如何使用 这里只介绍怎么在项目中安装引用和简单的 ...
- 匈牙利算法&模板O(mn)HDU2063
#include<cstdio> #include<cstring> #define maxn 510 using namespace std; int k,g,b,x,y,a ...
- 29 August
P1352 Bosses' Masquerade 树形DP模板. #include <cstdio> #include <algorithm> using namespace ...