牛客网暑期ACM多校训练营（第一场） - J Different Integers（线段数组or莫队)-优快云博客

本文提供了一种解决ACM竞赛中不同整数计数问题的有效方法，利用离线处理和线段树数据结构来快速计算指定区间内的不同整数数量。

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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 a ₁, a ₂, ..., a _n and q pairs of integers (l ₁, r ₁), (l ₂, r ₂), ..., (l _q, r _q), find count(l ₁, r ₁), count(l ₂, r ₂), ..., count(l _q, r _q) where count(i, j) is the number of different integers among a ₁, a ₂, ..., a _i, a _j, a _{j + 1}, ..., a _n.

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 a

₁

, a

₂

, ..., a

.
The i-th of the following q lines contains two integers l

 and r

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 ≤ 10

⁵


* 1 ≤ a

 ≤ n
* 1 ≤ l

, r

 ≤ n
* The number of test cases does not exceed 10.


骚操作:直接把数组*2 然后求1-L,R-N 就变成 求R-L+N之间的不同数的个数了；注意，主席树会TLE；要用线段数组+离线

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <map>
 
using namespace std;
const int maxn=300000+5;
map<int,int> mp;
int data[maxn];
int a[maxn];
int ans[200000+5];
struct node{
    int l,r,id;
    bool operator<(node t)const{
        return r<t.r;
    }
}q[200000+5];
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];
        }
        n=n*2;
 
 
        for(int i=0;i<m;i++){
            int x,y;
            scanf("%d%d",&x,&y);
            q[i].l=y;
            q[i].r=x+n/2;
            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;
}

也可以莫队加上读入挂

#include<bits/stdc++.h>
 
using namespace std;
int n,q,a[100005];
int L,R,ans;
struct node{
    int l,r,id;
};
node temp[100005];
int sum[1000005];
int anw[1000005];
int block[100005];
int read()
{
    char ch=' ';
    int ans=0;
    while(ch<'0' || ch>'9')
        ch=getchar();
    while(ch<='9' && ch>='0')
    {
        ans=ans*10+ch-'0';
        ch=getchar();
    }
    return ans;
}
int cmp(node a,node b){
    if(block[a.l]==block[b.l])return block[a.r]<block[b.r];
    return block[a.l]<block[b.l];
}
void add(int x){
    if(sum[x]==0)ans++;sum[x]++;
}
void del(int x){
    sum[x]--;if(sum[x]==0)ans--;
}
int main()
{
 
    ios::sync_with_stdio(false);
    while(~scanf("%d%d",&n,&q)){
        memset(sum,0,sizeof(sum));
        for(int i=1;i<=n;i++){
            a[i]=read();
            block[i]=i/sqrt(n);
        }
        for(int i=1;i<=q;i++){
            temp[i].l=read();
            temp[i].r=read();
            temp[i].id=i;
        }
        sort(temp+1,temp+1+q,cmp);
        L=0;R=n+1;ans=0;
        for(int i=1;i<=q;i++){
            while(L<temp[i].l)L++,add(a[L]);
            while(R>temp[i].r)R--,add(a[R]);
            while(L>temp[i].l)del(a[L]),L--;
            while(R<temp[i].r)del(a[R]),R++;
            anw[temp[i].id]=ans;
        }
        for(int i=1;i<=q;i++)printf("%d\n",anw[i]);
 
    }
    return 0;
}