luogu P4137 Rmq Problem / mex

传送门w

一道莫队

完全不会

yfl dalao表示我太菜了这道题贼简单

qaqqq

orz我对不起讲过莫队的yfl dalao

真的是纯纯纯莫队emmm

#include<cstdio>
#include<cmath>
#include<algorithm>
#define sev en
using namespace std;
#define N 200010

int n,m,a[N],s,num[N],ans[N],L = 1,R;
struct EDGE{
    int l,r,blo,id;
    bool operator < (const EDGE &x) const{
    return blo < x.blo || (blo == x.blo && r < x.r);
    }
}e[N];

void add(int x){
    num[a[x]]++;
    int i = s;
    while(num[i])
    i++;
    s = i;
}

void del(int x){
    num[a[x]]--;
    if(num[a[x]] == 0)
    s = min(s,a[x]);
}

void Mo(){
    for(int i = 1;i <= m;i++){
    int l = e[i].l,r = e[i].r;
    while(L > l)
        add(--L);
    while(R < r)
        add(++R);
    while(L < l)
        del(L++);
    while(R > r)
        del(R--);
    ans[e[i].id] = s;
    }
}

int main(){
    scanf("%d%d",&n,&m);
    int sq = sqrt(n);
    for(int i = 1;i <= n;i++)
    scanf("%d",&a[i]);
    for(int i = 1;i <= m;i++){
    int x,y;
    scanf("%d%d",&x,&y);
    e[i].l = x;
    e[i].r = y;
    e[i].id = i;
    e[i].blo = (x - 1) / sq + 1;
    }
    sort(e + 1,e + m + 1);
    Mo();
    for(int i = 1;i <= m;i++)
    printf("%d
",ans[i]);
    return 0;
}
像是板子一样的莫队

qaqqq

我现在博客怎么都这么水了

 

原文地址:https://www.cnblogs.com/sevenyuanluo/p/10765288.html