POJ 2352 Stars 【树状数组】

Stars

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 38952 Accepted: 16944

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.

恩，题目大意就是说，星星按纵坐标从小到大，问的是一个星星在它左下角的星星个数就为该星星的等级，最后让输出从0到n-1 级的星星的个数，这里树状数组直接求 x 之前坐标的星星个数而忽略纵坐标，这里我离散化了一下，因为数组的大小WA了两次，应该不离散化也可以

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#define maxn 32010using namespace std;int n,c[maxn],l[maxn];int x[maxn],y[maxn],r[maxn];int lowbit(int t){    return t&(t^(t-1));}void update(int x){    for(int i=x;i<=n;i+=lowbit(i))        c[i]++;}int sum(int x){    int ans=0;    for(int i=x;i>0;i-=lowbit(i))        ans+=c[i];    return ans;}int bsearch(int x,int ll,int rr){    while(ll<=rr)    {        int mid=(ll+rr)>>1;        if(r[mid]==x)            return mid;        if(r[mid]>x)            rr=mid-1;        else            ll=mid+1;    }    return 0;}int main(){    while(~scanf("%d",&n))    {        int k=0;        memset(c,0,sizeof(c));        memset(l,0,sizeof(l));        for(int i=0;i<n;++i)        {            scanf("%d%d",&x[i],&y[i]);            r[++k]=x[i];        }        sort(r+1,r+k+1);        int j=2;        for(int i=2;i<=k;++i)        {            if(r[i]!=r[i-1])                r[j++]=r[i];        }        for(int i=0;i<n;++i)        {            int xx=bsearch(x[i],1,j);            l[sum(xx)]++;            update(xx);        }        for(int i=0;i<n;++i)            printf("%d\n",l[i]);    }    return 0;}