牛客网暑期ACM多校训练营（第一场）.Different Integers（分块）-优快云博客

本文介绍了一个ACM竞赛题目，旨在寻找给定序列中特定区间内的不同整数数量。通过预处理和排序查询来优化算法效率，实现快速响应大量查询。

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链接：https://www.nowcoder.com/acm/contest/139/J
来源：牛客网

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 524288K，其他语言1048576K
64bit IO Format: %lld

题目描述

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

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.

#include<bits/stdc++.h>
using namespace std;
int block[100010],a[100010],out[100010],sum[100010];
int n,m,i,j,t,L,R,ans;
struct ask
{
    int l,r,id;
}q[100010];
 
bool cmp(ask x,ask y)
{
    if(block[x.l]!=block[y.l]) return block[x.l]<block[y.l];
    else return x.r<y.r;
}

void inc(int x)
{
    if(sum[x]==0)ans++;sum[x]++;
} 

void dec(int x)
{
    sum[x]--;if(sum[x]==0)ans--;
}
 
signed main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(sum,0,sizeof sum);
        for(i=1;i<=n;i++)scanf("%d",&a[i]),block[i]=i/sqrt(n);
        for(i=1;i<=m;i++)scanf("%d%d",&q[i].l,&q[i].r),q[i].id=i;
        sort(q+1,q+1+m,cmp);
        L=0,R=n+1;ans=0;
        for(i=1;i<=m;i++)
        {
            while(L<q[i].l)
                L++,inc(a[L]);
            while(R>q[i].r)
                R--,inc(a[R]);
            while(L>q[i].l)
                dec(a[L]),L--;
            while(R<q[i].r)
                dec(a[R]),R++;
            out[q[i].id]=ans;
        }
        for(i=1;i<=m;i++)printf("%d\n",out[i]);
    }
    return 0;
}