题目链接：http://codeforces.com/problemset/problem/670/C
Moscow is hosting a major international conference, which is attended by nscientists 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 aiis 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 bjis 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
Input
3 2 3 2 2 3 2 2 3
Output
2
Input
6 6 3 1 1 3 7 5 1 2 3 4 5 2 3 4 5 1
Output
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.
离散化：通俗的讲，“离散化”就是把无穷大集合中的若干个元素映射为有限集合以便于统计的方法。具体地说，假设问题涉及int范围内的n个整数 a[1]~a[n]，这n个整数可能重复，去重以后共有m个整数。我们要把每个整数 a[i] 用一个 1~m 之间的整数代替，并且保持大小顺序不变， 即如果 a[i] 小于（或等于、大于）a[j]，那么代替 a[i] 的整数也小于（或等于、小于）代替a[j]的整数。
分析：m 部电影和 n 个人最多涉及 2*m + n 种语言，我们把所有的电影和人涉及的语言放进一个数组，排序并离散化， 用一个 1~2*m+n 之间的整数代替每种语言。此时我们就可以用数组直接统计会上述每种语言的人的数量，从而选择满足题目要求的电影。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=2e5+5;
int n,m;
int a[maxn],b[maxn],c[maxn];
int sum[maxn*3];
int cnt,mm;
int arr[maxn*3],num[maxn*3];
void discrete(){//离散化
sort(arr+1,arr+cnt+1);
for(int i=1;i<=cnt;i++){
if(i==1||arr[i]!=arr[i-1])
num[++mm]=arr[i];
}
}
int query(int x){//二分查找x的位置
return lower_bound(num+1,num+mm+1,x)-num;
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++){//将所有电影和人涉及的语言放进一个数组，排序并离散化
scanf("%d",&a[i]);
arr[++cnt]=a[i];
}
scanf("%d",&m);
for(int i=1;i<=m;i++){
scanf("%d",&b[i]);
arr[++cnt]=b[i];
}
for(int i=1;i<=m;i++){
scanf("%d",&c[i]);
arr[++cnt]=c[i];
}
discrete();//离散化
for(int i=1;i<=n;i++){
int id=query(a[i]);//统计每种语言的人的数量
++sum[id];
}
int bmax=-1,cmax=-1,ans=0;
for(int i=1;i<=m;i++){//选择满足题目要求的电影
int x=query(b[i]);
int y=query(c[i]);
if(sum[x]>bmax){//优先考虑让很高兴的人最多
bmax=sum[x],cmax=sum[y];
ans=i;
}else {
if(sum[x]==bmax&&sum[y]>cmax){//如果答案不唯一、则在此前提下再让比较高兴的人最多
bmax=sum[x],cmax=sum[y];
ans=i;
}
}
}
printf("%d\n",ans);
return 0;
}
参考：算法竞赛进阶指南