POJ-2352.Stats(树状数组简单应用)

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 58255		Accepted: 24860

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

The first line of the input file contains a number of stars N (1<=N<=15000). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, 0<=X,Y<=32000). There can be only one star at one point of the plane. Stars are listed in ascending order of Y coordinate. Stars with equal Y coordinates are listed in ascending order of X coordinate. 

Output

The output should contain N lines, one number per line. The first line contains amount of stars of the level 0, the second does amount of stars of the level 1 and so on, the last line contains amount of stars of the level N-1.

Sample Input

5
1 1
5 1
7 1
3 3
5 5

Sample Output

1
2
1
1
0

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

Source

Ural Collegiate Programming Contest 1999

 

话不多说，都在代码里了..

/*
    读题很容易知道，每一颗star i 只需要那些满足 j.time < i.time and j.row <= i.row的star即可，依题意我们知道这刚好就是输入顺序。
    so 我们只需要选择那些在某个star之前放的并且row小于等于他的star。
    所以我们可以把i.row插入，即可查出在某个数j插入之前row.i < row.j的数量
 */
#include <cstdio>
using namespace std;

const int maxn = 32000 + 5;
int n, c[maxn], ans[maxn];

int lowbit(int x) {
    return x & (-x);
}

void add(int x, int d) {
    while(x <= maxn) {
        c[x] += d;
        x += lowbit(x);
    }
}

int query(int x) {
    int sum = 0;
    while(x > 0) {
        sum += c[x];
        x -= lowbit(x);
    }
    return sum;
}

int main() {
    int x, y;
    scanf("%d", &n);
    for(int i = 0; i < n; i ++) {
        scanf("%d %d", &x, &y);
        x += 1;
        ans[query(x)] ++;
        add(x, 1);
    }
    for(int i = 0; i < n; i ++) {
        printf("%d\n", ans[i]);
    }
    return 0;
}

 

 

 

转载于:https://www.cnblogs.com/bianjunting/p/11236368.html