BZOJ-3343 The magic of the leader + block (large block sorting and two points)

Portal: https://www.luogu.org/problemnew/show/P2801

Reference: http://hzwer.com/2784.html The idea is very clear;)

ps: I am in Luogu A, BZOJ needs permission;

The meaning of the question: How many numbers than K are there in the interval query;

Ideas: block, update and query violently on both sides, and use two-point counting for the middle query; for each update, if necessary, remember to re-sort (another array corresponding to the interval);

 

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;
typedef long long ll;
const int maxn = 1000009;
const int bmaxn = 1009;
ll a[maxn],b[maxn],add[bmaxn];
int n,m;

int belong[maxn];       
 int num,l[bmaxn],r[bmaxn];         // block number; left end of i block; right end of i block 
int block;                       // block size

void reset(int id)
{
    int le = l[id],ri = r[id];
    for(int i=le; i<=ri; i++)b[i] = a[i];
    sort (b +le, b+ri+ 1 );
}

void build()
{
    block = sqrt(n);
    num = n/block;if(n%block)num++;
    for(int i=1; i<=num; i++)
        l[i]=(i-1)*block+1,r[i] = i * block;
    r[num] = n;
    for(int i=1; i<=n; i++)
        belong[i]  = (i-1)/block + 1;
    for(int i=1; i <= num; i++)
        reset(i);
}

void update(int lx,int rx,ll val)
{
    if(belong[lx]==belong[rx])
    {
        for(int i=lx;i<=rx;i++)a[i]+=val;
        reset(belong[lx]);
    }
    else
    {
        int li = r[belong[lx]],ri = l[belong[rx]];
        for(int i = lx; i<=li; i++)a[i]+=val;
            reset(belong[lx]);
        for(int i = ri; i<=rx; i++)a[i]+=val;
            reset(belong[rx]);
        for(int i = belong[lx] + 1;i < belong[rx]; i++)add[i]+=val;
    }
}

ll query(int lx,int rx,ll k)
{
    ll res = 0;
    if(belong[lx]==belong[rx])
    {
        for(int i=lx;i<=rx;i++)if(a[i] >= k - add[belong[lx]])res++;
    }
    else 
    {
        int li = r[belong[lx]],ri = l[belong[rx]];
        // cout<<li<<" "<<ri<<endl;
        for(int i = lx; i <= li; i++) if(a[i] >= k-add[belong[lx]])res++;
        for(int i = ri; i <= rx; i++) if(a[i] >= k-add[belong[rx]])res++;
        for(int i = belong[lx] + 1; i<belong[rx]; i++)
        {
            int le = l[i],ri = r[i];
            while(le <= ri)
            {
                int mid = (le+ri)>>1;
                if(b[mid] < k - add[i])
                    le = mid + 1;
                else ri = mid - 1;
            }
            // printf("%d\n",le);
            // cout<<r[i]-le+1<<endl;
            res+=r[i] - le + 1;
        }
    }
    return res;
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1; i<=n; i++)
        scanf("%lld", &a[i]);
    build();
    for(int i=1; i<=m; i++)
    {
        char s[20];
        int x,y;
        ll v;
        scanf("%s%d%d%lld",s,&x,&y,&v);
        if(s[0]=='M')
        {
            update(x,y,v);
        }
        else
        {
            ll ans = query(x,y,v);
            printf("%lld\n",ans);
        }
    }
    return 0;
}