BZOJ 3809: Gty的二逼妹子序列 & 3236: [Ahoi2013]作业 [莫队]

题意:

询问区间权值在$[a,b]$范围内种类数和个数


 

莫队

权值分块维护种类数和个数$O(1)-O(\sqrt{N})$

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define pii pair<int, int>
#define MP make_pair
#define fir first
#define sec second
typedef long long ll;
const int N=1e5+5, M=1e6+5, BN=320, BS=320;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}

int n, Q, a[N], l, r, x, y;
struct _blo{int l, r;} b[BN];
int block, m, pos[N];
struct Block{
    inline void ini(){
        block=sqrt( n );
        m=(n-1)/block+1;
        for(int i=1; i<=n; i++) pos[i]=(i-1)/block+1;
        for(int i=1; i<=m; i++) b[i].l=(i-1)*block+1, b[i].r=i*block; b[m].r=n;
    }
    int kind[BS], c[N], sum[BS];
    inline void add(int v) {
        c[v]++; sum[pos[v]]++;
        if(c[v]==1) kind[pos[v]]++;
    }
    inline void del(int v) {
        c[v]--; sum[pos[v]]--;
        if(c[v]==0) kind[pos[v]]--;
    }
    pii que(int l,int r) {
        int pl=pos[l], pr=pos[r], ans1=0, ans2=0;
        if(pl==pr) for(int i=l; i<=r; i++) ans1+=c[i], ans2+= c[i]>0;
        else{
            for(int i=pl+1; i<pr; i++) ans1+=sum[i], ans2+= kind[i];
            for(int i=l; i<=b[pl].r; i++) ans1+=c[i], ans2+= c[i]>0;
            for(int i=b[pr].l; i<=r; i++) ans1+=c[i], ans2+= c[i]>0;
        }
        return MP(ans1, ans2);
    }
}B;

struct meow{
    int l, r, x, y, qid;
    bool operator <(const meow &a) const{return pos[l]==pos[a.l] ? r<a.r : pos[l]<pos[a.l];}
}q[M];
pii ans[M];
void modui(){
    int l=1, r=0;
    for(int i=1; i<=Q; i++){
        while(r<q[i].r) r++, B.add(a[r]);
        while(r>q[i].r) B.del(a[r]), r--;
        while(l<q[i].l) B.del(a[l]), l++;
        while(l>q[i].l) l--, B.add(a[l]);
        ans[q[i].qid]= B.que(q[i].x, q[i].y);
    }
}

int main() {
    freopen("in","r",stdin);
    n=read(); Q=read();
    for(int i=1; i<=n; i++) a[i]=read();
    for(int i=1; i<=Q; i++) l=read(), r=read(), x=read(), y=read(), q[i]=(meow){l, r, x, y, i};
    B.ini();
    sort(q+1, q+1+Q);
    modui();
    for(int i=1; i<=Q; i++) printf("%d %d\n",ans[i].fir, ans[i].sec);
}

 

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posted @ 2017-03-20 08:27  Candy?  阅读(198)  评论(0编辑  收藏  举报