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  • 莫队算法

      

    block=n/sqrt(m*2/3);

    求区间颜色个数:

    #include<bits/stdc++.h>
    using namespace std;
    //input by bxd
    #define rep(i,a,b) for(int i=(a);i<=(b);i++)
    #define repp(i,a,b) for(int i=(a);i>=(b);--i)
    #define RI(n) scanf("%d",&(n))
    #define RII(n,m) scanf("%d%d",&n,&m)
    #define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
    #define RS(s) scanf("%s",s);
    #define ll long long
    #define pb push_back
    #define CLR(A,v)  memset(A,v,sizeof A)
    //////////////////////////////////
    #define inf 0x3f3f3f3f
    #define lson l,m,pos<<1
    #define rson m+1,r,pos<<1|1
    const int N=1e5+6;
    
    ll a[N],sum[N];
    struct node
    {
        ll l,r,id;
    }s[N];
    ll anss[N],ans,block;
    
    bool cmp(node a,node b)
    {
        return (a.l/block)==(b.l/block)?a.r<b.r:a.l<b.l;
    }
    
    void del(int x)
    {
        sum[a[x]]--;
        if(sum[a[x]]==0)ans--;
    }
    void add(int x)
    {
        sum[a[x]]++;
        if(sum[a[x]]==1)ans++;
    }
    int n,m,k;
    int main()
    {
        cin>>n>>m>>k;
        block=n/sqrt(m*2/3);
    
        rep(i,1,n)RI(a[i]);
    
        rep(i,1,m)
            scanf("%lld%lld",&s[i].l,&s[i].r),s[i].id=i;
        
        sort(s+1,s+1+m,cmp);
        int l=1,r=0;//指针
        rep(i,1,m)
        {
            ll L=s[i].l,R=s[i].r;//范围
            while(l<L)del(l++);
            while(r>R)del(r--);
            while(l>L)add(--l);
            while(r<R)add(++r);
            anss[s[i].id]=ans;
        }
        rep(i,1,m)
        cout<<anss[i]<<endl;
    
        return 0;
    }
    View Code

     

    block=n/sqrt(m*2/3);

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  • 原文地址:https://www.cnblogs.com/bxd123/p/10917419.html
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