POJ 3468 A Simple Problem with Integers （分块）

最新推荐文章于 2020-10-30 22:12:42 发布

邵光亮

最新推荐文章于 2020-10-30 22:12:42 发布

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本文深入解析了一种基于分块思想的算法，该算法用于高效处理大规模数据集上的区间加法和查询操作，相较于线段树和树状数组，它提供了一种更易于理解的解决方案，尽管其复杂度稍高。

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Description

You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Outpu

Hint

The sums may exceed the range of 32-bit integers.

这次用的是分块的思想，比线段树好理解，比树状数组更好理解，但是复杂度高了。

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stdlib.h>
#include<queue>
#include<map>
#include<iomanip>
#include<math.h>
using namespace std;
typedef long long ll;
typedef double ld;
const int INF = 0x3f3f3f3f;
const int N=1000010;
ll a[N],sum[N],add[N];
int L[N],R[N];  //每一段的左右端点
int pos[N]; //每个位置属于哪一段
int n,m,t;

void change(int l,int r,ll d)
{
    int p = pos[l];
    int q = pos[r];
    if(p==q)
    {
        for(int i=l; i<=r; i++)
            a[i]+=d;
        sum[p] += d*(r-l+1);
    }
    else
    {
        for(int i=p+1; i<=q-1; i++)
            add[i]+=d;
        for(int i=l; i<=R[p]; i++)
            a[i]+=d;
        sum[p]+=d*(R[p]-l+1);
        for(int i=L[q]; i<=r; i++)
            a[i]+=d;
        sum[q]+=d*(r-L[q]+1);
    }
}

ll ask(int l,int r)
{
    int p = pos[l];
    int q = pos[r];
    ll ans=0;
    if(p==q)
    {
        for(int i=l; i<=r; i++)
            ans+=a[i];
        ans+=add[p]*(r-l+1);
    }
    else
    {
        for(int i=p+1; i<=q-1; i++)
            ans+=sum[i]+add[i]*(R[i]-L[i]+1);
        for(int i=l; i<=R[p]; i++)
            ans+=a[i];
        ans+=add[p]*(R[p]-l+1);
        for(int i=L[q]; i<=r; i++)
            ans+=a[i];
        ans += add[q]*(r-L[q]+1);
    }
    return ans;
}

int main()
{
    scanf("%d %d",&n,&m);
    for(int i=1; i<=n; i++)
        scanf("%lld",&a[i]);
    //分块
    t=sqrt(n);
    for(int i=1; i<=t; i++)
    {
        L[i]=(i-1)*sqrt(n)+1;
        R[i]=i*sqrt(n);
    }
    if(R[t]<n)
    {
        t++;
        L[t]=R[t-1]+1;
        R[t]=n;
    }
    //预处理
    for(int i=1; i<=t; i++)
    {
        for(int j=L[i]; j<=R[i]; j++)
        {
            pos[j]=i;
            sum[i]+=a[j];
        }
    }
    while(m--)
    {
        char op[3];
        int l,r,d;
        scanf("%s %d %d",op,&l,&r);
        if(op[0]=='C')
        {
            scanf("%d",&d);
            change(l,r,d);
        }
        else
            printf("%lld\n",ask(l,r));
    }
    return 0;
}