[解题报告]POJ2352 Stars 树状数组
刚学完树状数组,趁热打铁,AC了一道树状数组的基本题
因为x,y已经排序好,所以后输入的星星的不可能在先输入的星星左下,可以一边输入一边更新树状数组,c[i]表示的是树状元素中的值,sum(i)即可求出该位置左边一共有多少星星(先输入的星星一定不会在它上方,可以不考虑y),之后跟新这一点的值..需要注意的是,坐标有可能为0,而树状数组是从1开始的,所以对应的x+1再传入树状数组中,防止TLE.
/* Stars Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 17306 Accepted: 7519 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 */ #include <iostream> #include <cstdio> #include <string.h> using namespace std; int c[32001]; int a[15010]; int lowbit(int x){ return x&(-x); } void modify(int i){ while(i<=32001){ c[i]++; i+=lowbit(i); } } int sum(int i){ int s=0; while(i>0){ s+=c[i]; i-=lowbit(i); } return s; } int main(){ int n; scanf("%d",&n); int x,y; for(int i=1;i<=n;i++){ scanf("%d%d",&x,&y); a[sum(++x)]++;//防止0的出现,刚学树状数组,没有注意到这一点,会导致TLE modify(x); } for(int i=0;i<n;i++){ printf("%d/n",a[i]); } return 0; }