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.
这题我是用来练习树状数组的， 思路：因为输入是按照按照y递增，如果y相同则x递增的顺序给出的， 所以，对于第i颗星星，它的vis就是之前出现过的星星中，横坐标x小于等于i星横坐标的那些星星的总数量（前面的y一定比后面的y小）。 所以，需要找到一种数据结构来记录所有星星的x值，方便的求出所有值为0～x的星星总数量。 这题关键在于树状数组的运用
就是这段x为什么要加1，不加的话就超时了
for(int i = 1;i <= n;i++){ int x,y; scanf("%d %d",&x,&y); vis[sum(x+1)]++; add(x+1); }AC代码
#include <stdio.h> #include <algorithm> #define maxn 32005 using namespace std; int bit[32006]; int vis[32006]; void add(int i) { while(i <= maxn){ bit[i] += 1; i += i&(-i); } return; } int sum(int i) { int s = 0; while(i > 0){ s += bit[i]; i -= i&(-i); } return s; } int main() { int n; scanf("%d",&n); for(int i = 1;i <= n;i++){ int x,y; scanf("%d %d",&x,&y); vis[sum(x+1)]++; add(x+1); } for(int i = 0;i < n;i++){ printf("%d\n",vis[i]); } return 0; }