Stars（树状数组）

http://acm.hdu.edu.cn/showproblem.php?pid=1541

思路：对于两个点i,j,若xi>=xj,且yi>=yj,点i才能升一个level，现在对于后面输入的点的y值肯定比前面输入的点的y值要来的大，那么比较x就好了，现在就回到跟前面一篇的思路差不多的地方了，只是这里要求的是对于每一个出现的点，他的level的确定取决于前面有多少个比他小的x的点的个数。

Stars

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6619    Accepted Submission(s): 2640

Problem 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

51 15 17 13 35 5

 

Sample Output

12110

 

Source

Ural Collegiate Programming Contest 1999

 

#include <cstdio>#include <cstring>int n;int c[32010];int ans[15010];int x[15010];int bit(int t){    return t&(-t);}void update(int pos){    while(pos<=32010){        c[pos]++;        pos+=bit(pos);    }}int getsum(int pos){    int ans=0;    while(pos>0){        ans+=c[pos];        pos-=bit(pos);    }    return ans;}int main(){    while(scanf("%d",&n)!=EOF){        memset(c,0,sizeof(c));        memset(ans,0,sizeof(ans));        for(int i=0;i<n;i++){            int y;            scanf("%d%d",&x[i],&y);            x[i]++;            int k=getsum(x[i]);//k表示当前元素属于哪一个level的            ans[k]++;            update(x[i]);//出现的点的地方加一个1        }        for(int i=0;i<n;i++){                printf("%d\n",ans[i]);        }    }    return 0;}

AC之路，我选择坚持~~