知识概览
树状数组有两个作用：
- 快速求前缀和 时间复杂度O(log(n))
- 修改某一个数 时间复杂度O(log(n))
例题展示
- 单点修改，区间查询
题目链接
题解
涉及单点修改和求前缀和，并且要求时间复杂度小，可以用树状数组。
代码
cpp
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 200010;
int n;
int a[N];
int tr[N];
int Greater[N], lower[N];
int lowbit(int x)
{
return x & -x;
}
void add(int x, int c)
{
for (int i = x; i <= n; i += lowbit(i)) tr[i] += c;
}
int sum(int x)
{
int res = 0;
for (int i = x; i; i -= lowbit(i)) res += tr[i];
return res;
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
for (int i = 1; i <= n; i++)
{
int y = a[i];
Greater[i] = sum(n) - sum(y);
lower[i] = sum(y - 1);
add(y, 1); //将y加入树状数组，即数字y出现1次
}
memset(tr, 0, sizeof tr);
LL res1 = 0, res2 = 0;
for (int i = n; i; i--)
{
int y = a[i];
res1 += Greater[i] * (LL)(sum(n) - sum(y));
res2 += lower[i] * (LL)(sum(y - 1));
add(y, 1); //将y加入树状数组，即数字y出现1次
}
printf("%lld %lld\n", res1, res2);
return 0;
}2.区间修改，单点查询
题目链接
活动 - AcWing本活动组织刷《算法竞赛进阶指南》，系统学习各种编程算法。主要面向有一定编程基础的同学。
https://www.acwing.com/problem/content/248/
题解
需要用到差分数组，区间修改可以转化成对差分数组的单点修改，单点查询可以转化成对差分数组求前缀和，这样就可以转化成经典的树状数组操作。
代码
cpp
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 100010;
int n, m;
int a[N];
LL tr[N];
int lowbit(int x)
{
return x & -x;
}
void add(int x, int c)
{
for (int i = x; i <= n; i += lowbit(i)) tr[i] += c;
}
LL sum(int x)
{
LL res = 0;
for (int i = x; i; i -= lowbit(i)) res += tr[i];
return res;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
for (int i = 1; i <= n; i++) add(i, a[i] - a[i - 1]);
while (m--)
{
char op[2];
int l, r, d;
scanf("%s%d", op, &l);
if (*op == 'C')
{
scanf("%d%d", &r, &d);
add(l, d), add(r + 1, -d);
}
else
{
printf("%lld\n", sum(l));
}
}
return 0;
}参考资料
- AcWing算法提高课