Stars POJ2352（树状数组模板题）

Stars

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

Submit Status

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

Hint

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

题目分析：因为X,Y已经排好序了，对于这道题而言，可以不考虑Y，因为后面出现的点，Y肯定是大于前面的，只需考虑X就好，即：计算前面出现的X中有多少个小于当前的这个。然后树状数组敲敲模板就好。

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;const int maxx=32000+50;int c[maxx],res[maxx];int n;int lowbit(int x){    return x&-x;}void addnum(int x,int d){    while(x<maxx)    {        c[x]+=d;        x+=lowbit(x);    }}int getsum(int x){int ans=0;while(x>0){    ans+=c[x];    x-=lowbit(x);}return ans;}int main(){  scanf("%d",&n);int a,b;int ji=n;memset(c,0,sizeof(c));memset(res,0,sizeof(res));while(ji--){    scanf("%d%d",&a,&b);    addnum(a+1,1);      b=getsum(a+1);   res[b]++;}for(int i=1;i<=n;i++)printf("%d\n",res[i]);return 0;}