AcWing103 电影 (排序 + 离散化)

题目链接：https://www.acwing.com/problem/content/105/

首先对所有涉及到的语言离散化一下，统计会每种语言的人的个数
然后扫一遍电影，找到知道电影语言最多的人的那个电影

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
using namespace std;
const int maxn = 600010;
typedef long long ll;

int n, m, cnt, q;
int L[maxn], a[maxn], b[maxn], c[maxn], lan[maxn], con[maxn], id[maxn];

ll read(){ ll s = 0, f = 1ll; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1ll; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	n = read();
	for(int i=1;i<=n;++i) a[i] = read(), L[++cnt] = a[i];
	m = read();
	for(int i=1;i<=m;++i) b[i] = read(), L[++cnt] = b[i]; 
	for(int i=1;i<=m;++i) c[i] = read(), L[++cnt] = c[i]; 
	
	sort(L+1,L+1+cnt);
	
	for(int i=1;i<=cnt;++i){
		if(i == 1 || L[i] != L[i-1]){
			lan[++q] = L[i];
		}
	}
	
	for(int i=1;i<=m;++i){
		int pos = lower_bound(lan + 1,lan + 1 + q, b[i]) - lan; 
		id[pos] = i;
	}
	
	for(int i=1;i<=n;++i){
		int pos = lower_bound(lan + 1, lan + 1 + q, a[i]) - lan;
		++con[pos];
	} 
	
	int ans = 0, mxb = 0, mxc = 0;	

	for(int i=1;i<=m;++i){
		int posb = lower_bound(lan + 1, lan + 1 + q, b[i]) - lan;
		int posc = lower_bound(lan + 1, lan + 1 + q, c[i]) - lan;
		if(mxb < con[posb]){
			ans = i;
			mxb = con[posb];
			mxc = con[posc];
		}else if(mxb == con[posb]){
			if(mxc < con[posc]){
				ans = i;
				mxc = con[posc];
			}
		}
	} 
	printf("%d
",ans);
	
	return 0;
}