hdu4177:Super Mario
主席树+离散化。给一段区间。多次询问[l,r]中有多少个数小于k。啊主席树用指针版写出来优美多了QAQ。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define REP(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();bool f=true;
while(!isdigit(c)){
if(c=='-') f=false;c=getchar();
}
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return f?x:-x;
}
const int nmax=100005;
const int inf=0x7f7f7f7f;
struct node{
node *l,*r;int sum;
};
node *root[nmax],nodes[nmax*20],*pt;
int a[nmax],b[nmax];
node* build(int l,int r){
node* op=pt++;op->sum=0;
if(l==r) return op;
int mid=(l+r)>>1;
op->l=build(l,mid);op->r=build(mid+1,r);
return op;
}
node* update(int p,int add,node* t,int l,int r){
node* op=pt++;op->sum=t->sum+add;
if(l==r) return op;
int mid=(l+r)>>1;
if(p<=mid) op->r=t->r,op->l=update(p,add,t->l,l,mid);
else op->l=t->l,op->r=update(p,add,t->r,mid+1,r);
return op;
}
int query(node* t,node* s,int x,int l,int r){
if(l==r) return t->sum-s->sum;
int mid=(l+r)>>1;
if(x<=mid) return query(t->l,s->l,x,l,mid);
else return query(t->r,s->r,x,mid+1,r)+t->l->sum-s->l->sum;
}
int main(){
int cas=read();
REP(i,1,cas){
pt=nodes;
int n=read(),m=read();
REP(j,1,n) a[j]=b[j]=read();
sort(b+1,b+n+1);
int N=unique(b+1,b+n+1)-b-1;
N++;b[N]=inf; root[0]=build(1,N);
REP(j,1,n){
int k=lower_bound(b+1,b+N+1,a[j])-b;
root[j]=update(k,1,root[j-1],1,N);
} printf("Case %d:\n",i);
REP(j,1,m){
int l=read(),r=read(),v=read();l++,r++;
int k=upper_bound(b+1,b+1+N,v)-b-1;
int ans=0;
if(k>0) ans=query(root[r],root[l-1],k,1,N);
printf("%d\n",ans);
}
}
return 0;
}
Super Mario
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4662 Accepted Submission(s):
2148
amazing jumping ability reminded in our memory. Now the poor princess is in
trouble again and Mario needs to save his lover. We regard the road to the
boss’s castle as a line (the length is n), on every integer point i there is a
brick on height hi. Now the question is how many bricks in [L, R] Mario can hit
if the maximal height he can jump is H.
data.
For each test data:
The first line contains two integers n, m (1
<= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the
number of queries.
Next line contains n integers, the height of each brick,
the range is [0, 1000000000].
Next m lines, each line contains three integers
L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)
starting from 1) followed by m lines, each line contains an integer. The ith
integer is the number of bricks Mario can hit for the ith query.
10 10
0 5 2 7 5 4 3 8 7 7
2 8 6
3 5 0
1 3 1
1 9 4
0 1 0
3 5 5
5 5 1
4 6 3
1 5 7
5 7 3
4
0
0
3
1
2
0
1
5
1
hdu4177:Super Mario的更多相关文章
- HDU 4417 Super Mario(主席树求区间内的区间查询+离散化)
Super Mario Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tota ...
- Teaching Your Computer To Play Super Mario Bros. – A Fork of the Google DeepMind Atari Machine Learning Project
Teaching Your Computer To Play Super Mario Bros. – A Fork of the Google DeepMind Atari Machine Learn ...
- 主席树:HDU 4417 Super Mario
Super Mario Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total ...
- hdu4417 Super Mario 树阵离线/划分树
http://acm.hdu.edu.cn/showproblem.php?pid=4417 Super Mario Time Limit: 2000/1000 MS (Java/Others) ...
- Super Mario
Super Mario Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit ...
- hdu4417(Super Mario)—— 二分+划分树
Super Mario Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total ...
- hdu 4417 Super Mario 树状数组||主席树
Super Mario Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Prob ...
- HDU 4417 Super Mario(线段树)
Super Mario Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tota ...
- HDU 4417 Super Mario(划分树)
Super Mario Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total ...
随机推荐
- web一次请求的流程
1.客户端(浏览器输入网址)请求 2.发送http协议到web服务器(nginx),检测请求类别,如果时纯静态页面,则返响应返回给客户端. 3.如果有动态脚本(php语法)启动fastcgi进程,用解 ...
- 在制作joomla模板过程中遇到的问题
'''问题1.'''在jjc首页中两个通知公告和基建首页的两个模块中,当我点击查看文章标题是,而通知公告和最新动态页一直都还显示,发现文章一直在网站的下部,而不显示在它应该显示的main_rigth模 ...
- c# 代理IP获取通用方法
调用: ConcurrentQueue<string> proxyIpQueue = new ConcurrentQueue<string>(); Grab_ProxyIp(p ...
- 1079. Total Sales of Supply Chain (25)
时间限制 250 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A supply chain is a network of r ...
- Python socket模拟HTTP请求
0x00 起 最近在做一个对时间要求比较高的扫描器,需要封装一下SOCKET模拟HTTP发包的一些常用函数.简单的说,就是重写一下requests中的get.post方法. 今天在写的时候,遇到一枚很 ...
- Xcode文档下载与安装路径
https://developer.apple.com/library/downloads/docset-index.dvtdownloadableindex ~/Library/Developer/ ...
- 【BZOJ 1791】 [Ioi2008]Island 岛屿
Description 你将要游览一个有N个岛屿的公园.从每一个岛i出发,只建造一座桥.桥的长度以Li表示.公园内总共有N座桥.尽管每座桥由一个岛连到另一个岛,但每座桥均可以双向行走.同时,每一对这样 ...
- 关于Eigen库在Visual Studio2013中传参对齐报错问题
Error as follow: 具体问题及解决办法描述如下: (引自http://www.fx114.net/qa-278-97757.aspx) /************************ ...
- 获取网页上数据(图片、文字、视频)-b
Demo地址:http://download.csdn.net/detail/u012881779/8831835 获取网页上所有图片.获取所有html.获取网页title.获取网页内容文字... . ...
- Java 类加载机制 ClassLoader Class.forName 内存管理 垃圾回收GC
[转载] :http://my.oschina.net/rouchongzi/blog/171046 Java之类加载机制 类加载是Java程序运行的第一步,研究类的加载有助于了解JVM执行过程,并指 ...