牛客网暑期ACM多校训练营（第一场）J--Different Integers

Given a sequence of integers a1, a2, ..., an and q pairs of integers (l1, r1), (l2, r2), ..., (lq, rq), find count(l1, r1), count(l2, r2), ..., count(lq, rq) where count(i, j) is the number of different integers among a1, a2, ..., ai, aj, aj + 1, ..., an.

输入描述:

The input consists of several test cases and is terminated by end-of-file.
The first line of each test cases contains two integers n and q.
The second line contains n integers a1, a2, ..., an.
The i-th of the following q lines contains two integers li and ri.

输出描述:

For each test case, print q integers which denote the result.

示例1

输入

复制

3 2
1 2 1
1 2
1 3
4 1
1 2 3 4
1 3

输出

复制

2
1
3

备注:

* 1 ≤ n, q ≤ 105
* 1 ≤ ai ≤ n
* 1 ≤ li, ri ≤ n
* The number of test cases does not exceed 10.

题意：给定长度为n的数组，m次查询，每次查询的是[1,l]与[r,n]之间不同数字的个数。

树状数组+离散化

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <map>
#define lowbit(x) x&(-x)
using namespace std;
const int maxn=300000+5;
map<int,int> mp;
int data[maxn];
int a[maxn];
int ans[maxn];
struct node{
    int l,r,id;
    bool operator<(node t)const{
        return r<t.r;
    }
}q[maxn];
int sum(int i){
    int ans=0;
    while(i>0){
        ans+=data[i];
        i-=i&-i;
    }
    return ans;
}
void add(int i,int x){
    while(i<maxn){
        data[i]+=x;
        i+=i&-i;
    }
}
  
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        fill(data,data+n*2+2,0);
        mp.clear();
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            a[i+n]=a[i];
        }
        for(int i=0;i<m;i++){
            int x,y;
            scanf("%d%d",&x,&y);
            q[i].l=y;
            q[i].r=x+n;
            q[i].id=i;
        }
        sort(q,q+m);
        int pre=1;
        for(int i=0;i<m;i++){
            for(int j=pre;j<=q[i].r;j++){
                if(mp[a[j]]!=0){
                    add(mp[a[j]],-1);
                }
                add(j,1);
                mp[a[j]]=j;
            }
            pre=q[i].r+1;
            ans[q[i].id]=sum(q[i].r)-sum(q[i].l-1);
        }
        for(int i=0;i<m;i++){
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}