题意概括
区间加法,区间询问小于一个数的个数。
正题
对于每个块,除原数组之外用一个vector来有序地存储所有数。当区间加时,对于每个完整块维护共同加数,对于不完整的块直接暴力加上再重新排序。当询问时,对于每个完整块在vector中二分,对于不完整的,直接暴力计数。
代码
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
#define MAXN 50005
int n, m, a[MAXN], p[MAXN], b[500], mm;
vector<int> v[500];
int opt, l, r, c;
int EF( vector<int> vec, int x ){//自力更生,手打二分万岁QAQ
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 + 1;
}
int query( int l, int r, int c ){
int ans(0);
if ( p[l] == p[r] ){
for ( int i = l; i <= r; ++i )
if ( a[i] + b[p[l]] < c ) ans++;
return ans;
}
for ( int i = l; p[i] == p[l]; ++i )
if ( a[i] + b[p[i]] < c ) ans++;
for ( int i = r; p[i] == p[r]; --i )
if ( a[i] + b[p[i]] < c ) ans++;
for ( int i = p[l] + 1; i <= p[r] - 1; ++i )
ans += EF( v[i], c - 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]);//重排。实际上可以归并排序,或者要用时再临时排序,这里偷了懒QAQ
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 * c ) );
else Add( l, r, c );
}
return 0;
}
总结
比起其他算法老长老长的代码,分块算法的灵活在本题中得到体现QAQ
数列分块系列目录
数列分块入门2 <-
原文地址:https://www.cnblogs.com/louhancheng/p/10051148.html
时间: 2024-11-29 18:32:12