CSU-ACM2017暑期训练16-树状数组 B

B - Stars

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

51 15 17 13 35 5

Sample Output

12110

Hint

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

题目需要求出每颗星左下方的星的数量。那么从左至右，从下至上地读取星的横坐标即可。因为在按照从下至上的顺序读取时已经默认不会取到当前位置上侧的数据了，这时只有横坐标对结果有贡献。所以只需读出横坐标小于当前位置的星的数量，将这个数量就是当前星的等级，将这个等级的数量加一，再将当前横坐标的星的数量加一。重复这个过程即得结果。

注意到这题的输入样例已经是排好序的了，所以直接在输入期间进行树状数组更新，不必操心排序的问题了。

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <queue>using namespace std;const int maxn = 32004;int n;int C[maxn], level[maxn];void add(int x, int d){    while(x < 32004){        C[x] += d; x += x&-x;    }}int sum(int x){    int ret = 0;    while(x > 0){        ret += C[x]; x -= x&-x;    }    return ret;}int main(){#ifdef TESTfreopen("test.txt", "r", stdin);#endif // TEST    int x, y;    while(scanf("%d", &n) != EOF){        memset(C, 0, sizeof(C));        memset(level, 0, sizeof(level));        for(int i = 0; i < n; i++){            scanf("%d%d", &x, &y);            level[sum(x+1)]++;            add(x+1, 1);        }        for(int i = 0; i < n; i++)            printf("%d\n", level[i]);    }    return 0;}