2018牛客网暑期ACM多校训练营(第一场) J - Different Integers - [莫队算法]
题目链接:https://www.nowcoder.com/acm/contest/139/J
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.
输入
3 2 1 2 1 1 2 1 3 4 1 1 2 3 4 1 3
输出
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:
给出序列a[1:n],给出q个查询,每个查询包含(l,r),要求回答a[1:l]和a[r:n]合起来有多少不同的整数。
题解:
莫队算法。
统计a[1:n]有多少不同的整数,记为sum,所有 r - l ≤ 1 的查询答案为sum,
cnt[k]表示整数k出现了几次,sum的变动根据cnt[k]的0→1以及1→0变化判断。
AC代码:
#include<bits/stdc++.h> using namespace std; const int MAXN=1e5+10; const int MAXQ=1e5+10; struct Query { int block; int id; int l,r; bool operator <(const Query &oth) const { return (this->block == oth.block) ? (this->r < oth.r) : (this->block < oth.block); } }query[MAXQ]; int n,q; int sum; int num[MAXN],cnt[MAXN]; int ans[MAXQ]; int main() { while(scanf("%d%d",&n,&q)!=EOF) { memset(cnt,0,sizeof(cnt)); sum=0; for(int i=1;i<=n;i++) { scanf("%d",&num[i]); if(cnt[num[i]]==0) sum++; cnt[num[i]]++; } int len=sqrt(n); for(int i=1;i<=q;i++) { scanf("%d%d",&query[i].l,&query[i].r); if(query[i].r-query[i].l<=1) ans[i]=sum; query[i].block=query[i].l/len; query[i].id=i; } sort(query+1,query+q+1); int pl=1,pr=2; for(int i=1;i<=q;i++) { if(query[i].r-query[i].l<=1) continue; if(pr<query[i].r) { for(int j=pr;j<query[i].r;j++) { cnt[num[j]]--; if(cnt[num[j]]==0) sum--; } } if(pr>query[i].r) { for(int j=pr-1;j>=query[i].r;j--) { if(cnt[num[j]]==0) sum++; cnt[num[j]]++; } } if(pl<query[i].l) { for(int j=pl+1;j<=query[i].l;j++) { if(cnt[num[j]]==0) sum++; cnt[num[j]]++; } } if(pl>query[i].l) { for(int j=pl;j>query[i].l;j--) { cnt[num[j]]--; if(cnt[num[j]]==0) sum--; } } ans[query[i].id]=sum; pl=query[i].l; pr=query[i].r; } for(int i=1;i<=q;i++) { printf("%d\n",ans[i]); } } }
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