A Simple Problem with Integers 线段树区间更新区间查询

最新推荐文章于 2021-01-24 14:10:20 发布

转载最新推荐文章于 2021-01-24 14:10:20 发布 · 92 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/joeylee97/p/7339582.html

本文介绍了一种使用线段树进行区间更新和查询的高效算法。通过实例演示了如何构建线段树，并实现区间加法操作和区间求和查询，适用于解决大规模数据集上的动态区间操作问题。

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 115624		Accepted: 35897
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a₊₁, ... , A_b.

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 Output

Hint

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

Source

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 100009
#define L 31
#define INF 1000000009
#define eps 0.00000001
/*
线段树 区间更新区间查询
*/
LL a[MAXN],pre[MAXN];
struct node
{
    LL l, r;
    LL data, sum, laz;
}T[MAXN*4];
void build(LL p, LL l, LL r)
{
    T[p].data = T[p].sum = T[p].laz = 0;
    T[p].l = l, T[p].r = r;
    if (l == r) return;
    LL mid = (l + r) / 2;
    build(p * 2, l, mid);
    build(p * 2 + 1, mid + 1, r);
}
void update(LL p, LL l, LL r, LL v)
{
    //cout << p << ' ' << l << ' ' << r << ' ' << v << endl;
    if (T[p].l >= l && T[p].r <= r)
    {
        T[p].data += v;
        T[p].laz = 1;
        T[p].sum += (T[p].r - T[p].l + 1) * v;
        return;
    }
    LL mid = (T[p].l + T[p].r) / 2;
    if (T[p].laz)
    {
        T[p].laz = 0;
        update(p * 2, T[p].l, mid, T[p].data);
        update(p * 2 + 1, mid + 1, T[p].r, T[p].data);
        T[p].data = 0;
    }
    if (r <= mid)
        update(p * 2, l, r, v);
    else if (l > mid)
        update(p * 2 + 1, l, r, v);
    else
    {
        update(p * 2, l, mid, v);
        update(p * 2 + 1, mid + 1, r, v);
    }
    T[p].sum = T[p * 2].sum + T[p * 2 + 1].sum;
}
LL query(LL p, LL l, LL r)
{
    if (l == T[p].l&&r == T[p].r)
        return T[p].sum;
    LL mid = (T[p].l + T[p].r) / 2;
    if (T[p].laz)
    {
        T[p].laz = 0;
        update(p * 2, T[p].l, mid, T[p].data);
        update(p * 2 + 1, mid + 1, T[p].r, T[p].data);
        T[p].data = 0;
    }
    if (r <= mid)
        return query(p * 2, l, r);
    else if (l > mid)
        return query(p * 2 + 1, l, r);
    else
        return query(p * 2, l, mid) + query(p * 2 + 1, mid + 1, r);
}
LL n, q;
int main()
{
    scanf("%lld%lld", &n, &q);
    for (LL i = 1; i <= n; i++)
        scanf("%lld", &a[i]), pre[i] = pre[i - 1] + a[i];
    char c[2];
    LL a, b, d;
    build(1,1,n);
    while (q--)
    {
        scanf("%s", c);
        if (c[0] == 'Q')
            scanf("%lld%lld", &a, &b), printf("%lld\n", query(1, a, b) + pre[b] - pre[a-1]);
        else
            scanf("%lld%lld%lld", &a, &b, &d), update(1, a, b, d);
    }
}

转载于:https://www.cnblogs.com/joeylee97/p/7339582.html

A Simple Problem with Integers 线段树 区间更新 区间查询

A Simple Problem with Integers 线段树区间更新区间查询