【洛谷】 SP3267 DQUERY - D-query 莫队算法模板题

题意:若干询问,每次询问输出【L,R】里面不同数的个数

思路:

莫队算法模板题了,记录一下

AC代码:

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e6+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline int read(){ int f = 1; int x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

int n;
int pos[maxn];
int a[maxn];
int cnt[maxn*10];
int Ans[maxn];
int sum = 0;

typedef struct Pos
{
    int l;
    int r;
    int id;
}P;
P q[maxn];

bool cmp(P a, P b)
{
    if(pos[a.l] == pos[b.l])
    {
        if(pos[a.l] & 1) return pos[a.r] < pos[b.r];
        return pos[a.r] > pos[b.r];
    }
    return pos[a.l] < pos[b.l];
}

void add(int x)
{
    if(!cnt[a[x]])  sum++;
    cnt[a[x]] ++;
}

void del(int x)
{
    cnt[a[x]] --;
    if(!cnt[a[x]]) sum --;
}

int main()
{
    n = read();
    int block = sqrt(n*1.0);
    rep(i,1,n) a[i] = read(), pos[i] = (i-1)/block + 1;
    int m = read();
    rep(i,1,m)
    {
        q[i].l = read(), q[i].r = read(), q[i].id = i;
    }
    sort(q+1,q+1+m,cmp);
    int L = 0, R = 0;
    rep(i,1,m)
    {
        while(q[i].l < L) add(--L);
        while(q[i].r > R) add(++R);
        while(q[i].l > L) del(L++);
        while(q[i].r < R) del(R--);
        Ans[q[i].id] = sum;
    }
    rep(i,1,m) cout<<Ans[i]<<'\n';
    return 0;
}

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转载自blog.csdn.net/qq_45492531/article/details/107804005