「分塊系列」數列分塊入門3 解題報告

時間 2020-01-08

標籤分塊系列數列入門解題報告欄目應用數學简体版

原文原文鏈接

數列分塊入門3

題意歸納

區間加法，區間求前驅。html

寫在前面

這題的方法與分塊2方法極其相似，建議自行解決。spa

正題

和上一題相似，可是二分不是用來計數的，而是用來求小於c的最大值的。而後對於不完整快，將小於c的值求最大值，再與全部塊中二分結果求最大值便可。(其餘思路上一篇題解已經講了，這裏再也不復述，代碼註釋也懶得打了，由於比較簡單，很容易理解)code

代碼

#include<cstdio>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
#define MAXN 100005

int n, m, a[MAXN], p[MAXN], b[1005], mm;
vector<int> v[1005];
int opt, l, r, c;

int EF( vector<int> vec, int x ){
    int l, r, mid, ans(-1);
    l = 0; r = vec.size() - 1;
    while( l <= r ){
        mid = ( l + r ) >> 1;
        if ( vec[mid] < x ){
            ans = mid;
            l = mid + 1;
        }
        else r = mid - 1;
    }
    return ans;
}

int query( int l, int r, int c ){
    int ans(-1);
    
    if ( p[l] == p[r] ){
        for ( int i = l; i <= r; ++i )
            if ( a[i] + b[p[l]] < c ) ans = max( ans, a[i] + b[p[l]] );
        return ans;
    }
    
    for ( int i = l; p[i] == p[l]; ++i )
        if ( a[i] + b[p[i]] < c ) ans = max( ans, a[i] + b[p[l]] );
    for ( int i = r; p[i] == p[r]; --i )
        if ( a[i] + b[p[i]] < c ) ans = max( ans, a[i] + b[p[r]] );
    for ( int i = p[l] + 1; i <= p[r] - 1; ++i ){
        int t(EF( v[i], c - b[i] ));
        if ( t >= 0 ) ans = max( ans, v[i][t] + b[i] );
    }
    return ans;
}

void re( int x ){
    v[x].clear();
    int be(( x - 1 ) * m + 1);
    for ( int i = be; p[i] == p[be]; i++ ) v[x].push_back( a[i] );
    sort( v[x].begin(), v[x].end() );
}

void Add( int l, int r, int c ){
    if ( p[l] == p[r] ){
        for ( int i = l; i <= r; ++i ) a[i] += c;
        re( p[l] ); return; 
    }
    
    for ( int i = l; p[i] == p[l]; ++i ) a[i] += c;
    re(p[l]);
    for ( int i = r; p[i] == p[r]; --i ) a[i] += c;
    re(p[r]);
    for ( int i = p[l] + 1; i < p[r]; ++i ) b[i] += c;
}

int main(){
    scanf( "%d", &n ); m = (int)sqrt(n);
    
    for ( int i = 1; i <= n; ++i ) p[i] = ( i - 1 ) / m + 1, mm = p[i];
    for ( int i = 1; i <= n; ++i ) scanf( "%d", &a[i] );
    for ( int i = 1; i <= n; ++i ) v[p[i]].push_back(a[i]);
    for ( int i = 1; i <= mm; ++i ) sort( v[i].begin(), v[i].end() );
    
    for ( int i = 1; i <= n; ++i ){
        scanf( "%d%d%d%d", &opt, &l, &r, &c );
        if ( opt ) printf( "%d\n", query( l, r, c ) );
        else Add( l, r, c );
    }
    return 0;
}