Cinema CodeForces - 670C (离散+排序)
Moscow is hosting a major international conference, which is attended by n scientists from different countries. Each of the scientists knows exactly one language. For convenience, we enumerate all languages of the world with integers from 1 to 109.
In the evening after the conference, all n scientists decided to go to the cinema. There are m movies in the cinema they came to. Each of the movies is characterized by two distinct numbers — the index of audio language and the index of subtitles language. The scientist, who came to the movie, will be very pleased if he knows the audio language of the movie, will be almost satisfied if he knows the language of subtitles and will be not satisfied if he does not know neither one nor the other (note that the audio language and the subtitles language for each movie are always different).
Scientists decided to go together to the same movie. You have to help them choose the movie, such that the number of very pleased scientists is maximum possible. If there are several such movies, select among them one that will maximize the number of almost satisfied scientists.
Input
The first line of the input contains a positive integer n (1 ≤ n ≤ 200 000) — the number of scientists.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the index of a language, which the i-th scientist knows.
The third line contains a positive integer m (1 ≤ m ≤ 200 000) — the number of movies in the cinema.
The fourth line contains m positive integers b1, b2, ..., bm (1 ≤ bj ≤ 109), where bj is the index of the audio language of the j-th movie.
The fifth line contains m positive integers c1, c2, ..., cm (1 ≤ cj ≤ 109), where cj is the index of subtitles language of the j-th movie.
It is guaranteed that audio languages and subtitles language are different for each movie, that is bj ≠ cj.
Output
Print the single integer — the index of a movie to which scientists should go. After viewing this movie the number of very pleased scientists should be maximum possible. If in the cinema there are several such movies, you need to choose among them one, after viewing which there will be the maximum possible number of almost satisfied scientists.
If there are several possible answers print any of them.
Examples
3
2 3 2
2
3 2
2 3
2
6
6 3 1 1 3 7
5
1 2 3 4 5
2 3 4 5 1
1
Note
In the first sample, scientists must go to the movie with the index 2, as in such case the 1-th and the 3-rd scientists will be very pleased and the 2-nd scientist will be almost satisfied.
In the second test case scientists can go either to the movie with the index 1 or the index 3. After viewing any of these movies exactly two scientists will be very pleased and all the others will be not satisfied.
题意:有n个科学家,每个人都会一种语言,用整数a【i】表示,m部电影,每个电影用b【i】语言配音,用c【i】语言填字幕,
求一个电影,他被最多的人听懂,如果有多种方案,求此情况下,被最多人看懂的(a【i】 == c【i】就可以看懂)。
思路: 我们可以想到,把科学家会的语言计数,然后对电影先按b【i】数量后按c【i】数量排序。
计数的话,我们肯定要离散化,因为 0 <= i <= 2e5
1、我们可以用map,但是注意这题是卡了unordered_map的,直接用map或者hash_map都行
include<bits/stdc++.h>
using namespace std; int n,m;
const int maxn = 2e5+;
int sc[maxn]; map<int,int>mp;
struct Mo
{
int b;
int c;
int id;
} mov[maxn]; bool cmp(Mo a,Mo b)
{
if(a.b == b.b)
return a.c > b.c;
return a.b > b.b;
}
int main()
{
scanf("%d",&n);
int tmp;
int cnt = ;
for(int i=; i<=n; i++)
{
scanf("%d",&tmp);
if(!mp[tmp])
{
mp[tmp] = ++cnt;
sc[cnt]++;
}
else
sc[mp[tmp]]++;
}
scanf("%d",&m);
for(int i=; i<=m; i++)
scanf("%d",&tmp),mov[i].id = i,mov[i].b = sc[mp[tmp]];
for(int i=; i<=m; i++)
scanf("%d",&tmp),mov[i].c = sc[mp[tmp]];
sort(mov+,mov++m,cmp);
printf("%d\n",mov[].id);
}
2、我们可以用二分离散,就是先排序a【i】,然后用unique或者手写去重,用二分查找a【i】在去重后的哪个位置,得出离散后的结果
这样我们在求取b【i】数量(或c【i】)的时候,也是二分查找b【i】(或c【i】)在去重后的位置,但是的出来位置上的电影编号不一定等于b【i】(或c【i】),
如果相等那么b【i】 = sc【pos】 (sc是之前科学家语言的计数),如果不等,b【i】 = 0(说明科学家中没人懂这语言);
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std; int n,m;
const int maxn = 2e5+;
int a[maxn];
int b[maxn];
int sc[maxn]; struct Mo
{
int b;
int c;
int id;
} mov[maxn]; bool cmp(Mo a,Mo b)
{
if(a.b == b.b)
return a.c > b.c;
return a.b > b.b;
}
int query(int x,int len)
{
return lower_bound(b+,b++len,x)-b;
}
int main()
{
scanf("%d",&n);
for(int i=; i<=n; i++)scanf("%d",&a[i]);
sort(a+,a++n);
int tmp;
int len = ;
for(int i=;i<=n;i++)
{
if(i== || a[i] != a[i-])
b[++len] = a[i];
} for(int i=;i<=n;i++)
{
tmp = query(a[i],len);
sc[tmp]++;
}
scanf("%d",&m);
for(int i=; i<=m; i++)
{
scanf("%d",&tmp),mov[i].id = i;
int pos = query(tmp,len);
if(b[pos] == tmp)mov[i].b = sc[pos];
else mov[i].b = ;
}
for(int i=; i<=m; i++)
{
scanf("%d",&tmp);
int pos = query(tmp,len);
if(b[pos] == tmp)mov[i].c = sc[pos];
else mov[i].c = ;
}
sort(mov+,mov++m,cmp);
printf("%d\n",mov[].id);
}
Cinema CodeForces - 670C (离散+排序)的更多相关文章
- CodeForces 670C Cinema(排序,离散化)
C. Cinema time limit per test 2 seconds memory limit per test 256 megabytes input standard input out ...
- 【Codeforces 670C】 Cinema
[题目链接] http://codeforces.com/contest/670/problem/C [算法] 离散化 [代码] #include<bits/stdc++.h> using ...
- CodeForces 670C Cinema
简单题. 统计一下懂每种语言的人分别有几个,然后$O(n)$扫一遍电影就可以得到答案了. #pragma comment(linker, "/STACK:1024000000,1024000 ...
- CodeForces 558E(计数排序+线段树优化)
题意:一个长度为n的字符串(只包含26个小字母)有q次操作 对于每次操作 给一个区间 和k k为1把该区间的字符不降序排序 k为0把该区间的字符不升序排序 求q次操作后所得字符串 思路: 该题数据规模 ...
- Division and Union CodeForces - 1101C (排序后处理)
There are nn segments [li,ri][li,ri] for 1≤i≤n1≤i≤n. You should divide all segments into two non-emp ...
- Codeforces 997D(STL+排序)
D. Divide by three, multiply by two time limit per test 1 second memory limit per test 256 megabytes ...
- CF思维联系--CodeForces -214C (拓扑排序+思维+贪心)
ACM思维题训练集合 Furik and Rubik love playing computer games. Furik has recently found a new game that gre ...
- CodeForces - 721C 拓扑排序+dp
题意: n个点m条边的图,起点为1,终点为n,每一条单向边输入格式为: a,b,c //从a点到b点耗时为c 题目问你最多从起点1到终点n能经过多少个不同的点,且总耗时小于等于t 题解: 这道 ...
- CodeForces 22D Segments 排序水问题
主题链接:点击打开链接 升序右键点.采取正确的点 删边暴力 #include <cstdio> #include <cstring> #include <algorith ...
随机推荐
- Oracle 之 外部表
一.外部表概述 外部表只能在Oracle 9i 之后来使用.简单地说,外部表,是指不存在于数据库中的表. 通过向Oracle 提供描述外部表的元数据,我们可以把一个操作系统文件当成一个只读的数 据库表 ...
- eclipse 安装教程
eclipse 安装教程 一:安装包下载: 链接: https://pan.baidu.com/s/1qZtt62o 密码: 4ak2 注:若 下载链接失效,请看本文公告的QQ群,请联系群主. 二:安 ...
- Confluence 6 数据模型
本文档提供了 Confluence 的数据结构视图(schema )和数据模型概念上的的概述. 备注: Hibernate 的映射文件是针对 Confluence 数据模型的直接描述.在系统中的 Co ...
- Redis持久化概念
redis持久化概念 Author:SimpleWu GitHub-redis 什么是持久化? 概念:把内存的数据保存在磁盘的过程. Redis的持久化? redis是内存数据库,它把数据存储在内存中 ...
- LeetCode(110):平衡二叉树
Easy! 题目描述: 给定一个二叉树,判断它是否是高度平衡的二叉树. 本题中,一棵高度平衡二叉树定义为: 一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过1. 示例 1: 给定二叉树 [3, ...
- PyCharm新建.py文件时自动带出指定内容
如:给Pycharm加上头行 # coding:utf-8File—Setting—Editor--Code Style--File and Code Templates--Python Scrip ...
- 统计nginx日志里访问次数最多的前十个IP
awk '{print $1}' /var/log/nginx/access.log | sort | uniq -c | sort -nr -k1 | head -n 10
- C++ Primer 笔记——标准库类型string
1.如果使用等号初始化一个变量,实际上执行的是拷贝初始化,编译器吧等号右侧的初始值拷贝到新创建的对象中去:如果不使用等号则执行的是直接初始化. std::string str = "Test ...
- 即时通讯协议之XMPP
目前IM即时通信有四种协议 1.即时信息和空间协议(IMPP) 2.空间和即时信息协议(PRIM) 3.针对即时通讯和空间平衡扩充的进程开始协议SIP 4.XMPP协议: 该协议的前身是Jabber, ...
- UI开发总结
1. bootstrap tooltip 修改内容 <i class="ace-icon fa fa-user" id="test-tooltip" ti ...