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Language:Default
Stars
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 |
题意：题目会给n个（x，y），并且按照以下顺序依次给出数据：先按y递增，y相同按x递增。要我们给出0~n-1等级的个数，其中等级这样定义：有一个（xi，yi），等级为所有x<=xi && y<=yi（不包括本身）的个数。
思路：树状数组题，因为题目给出的y是递增的，所以我们只要关心当前（x，y）之前小于等于x的个数有多少，就是它的等级了。因为x>=0，我们用的树状数组所以要++。
代码：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<cmath>
#include<string>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<iostream>
#include<algorithm>
#include<sstream>
#define ll long long
const int N=32005;
const int INF=1e9;
using namespace std;
int a[N],ans[15005];
int n;
int lowbit(int x){
return x&(-x);
}
void update(int x){
for(int i=x;i<N;i+=lowbit(i)){
a[i]++;
}
}
int sum(int x){
int ans=0;
for(int i=x;i>=1;i-=lowbit(i)){
ans+=a[i];
}
return ans;
}
int main(){
int cnt,x,y;
while(~scanf("%d",&n)){
memset(a,0,sizeof(a));
memset(ans,0,sizeof(ans));
for(int i=0;i<n;i++){
scanf("%d%d",&x,&y);
ans[sum(++x)]++;
update(x);
}
for(int i=0;i<n;i++){
printf("%d\n",ans[i]);
}
}
return 0;
}