【POJ3468】 A Simple Problem with Integers （线段树）

最新推荐文章于 2020-11-24 12:54:48 发布

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本文详细介绍了一种高效的数据结构——线段树，在处理区间加法和区间求和操作的应用。通过具体实例，展示了线段树如何实现快速的区间修改和查询，特别适用于大量更新和查询操作的场景。

题目

描述

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.

输入

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.

输出

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

样例输入

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

样例输出

4
55
9
15

提示

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

思路

线段树，very easy。
~~然而半个多月没打线段树，错了n遍。~~
线段树的区间查询+区间修改。
输入错了n次（用了getchar）。
很简单。
上代码。

代码

#include <cstdio>
#include <iostream>
using namespace std;
int n, a, b, m, y, ans;
struct node {
    int l, r, w, f;
} tr[800001];
void build(int k, int ll, int rr) {
    tr[k].l = ll, tr[k].r = rr;
    if (tr[k].l == tr[k].r) {
        scanf("%d", &tr[k].w);
        return;
    }
    int m = (ll+rr)/2;
    build(k*2, ll, m);
    build(k*2+1, m+1, rr);
    tr[k].w = tr[k*2].w+tr[k*2+1].w;
}
inline void down(int k) {
    tr[k*2].f += tr[k].f;
    tr[k*2+1].f += tr[k].f;
    tr[k*2].w += tr[k].f*(tr[k*2].r-tr[k*2].l+1);
    tr[k*2+1].w += tr[k].f*(tr[k*2+1].r-tr[k*2+1].l+1);
    tr[k].f = 0;
}
void ask_interval(int k) {
    if (tr[k].l >= a && tr[k].r <= b) {
        ans += tr[k].w;
        return;
    }
    if (tr[k].f) down(k);
    int m = (tr[k].l+tr[k].r)/2;
    if (a <= m) ask_interval(k*2);
    if (b > m) ask_interval(k*2+1);
}
inline void change_interval(int k) {
    if (tr[k].l >= a && tr[k].r <= b) {
        tr[k].w += (tr[k].r-tr[k].l+1)*y;
        tr[k].f += y;
        return;
    }
    if (tr[k].f) down(k);
    int m = (tr[k].l+tr[k].r)/2;
    if (a <= m) change_interval(k*2);
    if (b > m) change_interval(k*2+1);
    tr[k].w = tr[k*2].w+tr[k*2+1].w;
}
int main() {
    scanf("%d%d", &n, &m);
    build(1, 1, n);
    for (int i = 1; i <= m; i++) {
        char p[5]; ans = 0; scanf("%s", &p);
        if (p[0] == 'Q') {
            scanf("%d%d", &a, &b);
            ask_interval(1);
            printf("%d\n", ans);
        }
        else {
             scanf("%d%d%d", &a, &b, &y);
             change_interval(1);
        }
    }
    return 0;
}