ACM模板-树状数组

一、模板

const int MAX = 1000010;
int c[MAX] = {0};
int lowbit(const int& x){ return (x)&(-x); }
void zero() { memset(c, 0, sizeof(c)); }
// 一定要小心零的问题
void modify(int i, const int& d)
{
    while (i <= MAX)
    {
        c[i] += d;
        i += lowbit(i);
    }
}
int sum(int i)
{
    int ret = 0;
    while (i) {
        ret += c[i];
        i -= lowbit(i);
    }
    return ret;
}

二、例题（POJ2352）

Description

Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers want to know the distribution of the levels of the stars.

For example, look at the map shown on the figure above. Level of the star number 5 is equal to 3 (it's formed by three stars with a numbers 1, 2 and 4). And the levels of the stars numbered by 2 and 4 are 1. At this map there are only one star of the level 0, two stars of the level 1, one star of the level 2, and one star of the level 3.You are to write a program that will count the amounts of the stars of each level on a given map.

Input

Output

Sample Input

5
1 1
5 1
7 1
3 3
5 5

Sample Output

1
2
1
1
0

Solution

#define _CRT_SECURE_NO_WARNINGS
#include "cstdio"
#include "iostream"
#include "algorithm"
#include "cstring"

using namespace std;
const int MAX = 1000010;
int c[MAX] = {0};
int lowbit(const int& x){ return (x)&(-x); }
void zero() { memset(c, 0, sizeof(c)); }
// 一定要小心零的问题
void modify(int i, const int& d)
{
    while (i <= MAX)
    {
        c[i] += d;
        i += lowbit(i);
    }
}
int sum(int i)
{
    int ret = 0;
    while (i) {
        ret += c[i];
        i -= lowbit(i);
    }
    return ret;
}

int R[MAX] = { 0 };
int main()
{
    int N,x,y;
    scanf("%d",&N);
    for (int i = 0; i < N; ++i) {
        scanf("%d%d", &x, &y); ++x;
        ++R[sum(x)];
        modify(x, 1);
    }
    for (int i = 0; i < N; ++i) {
        printf("%d\n",R[i]);
    }
    return 0;
}