HDU 1541 Stars 【线段树】

Stars

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6638    Accepted Submission(s): 2647

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

题目大意就是说，每个星星左边和下边的星星数为该星星的等级，然后问的是每个等级有几个星星。题目中输入信息是就是按照纵坐标从小到大，然后纵坐标相同的横坐标从小到大，所以未输入的星星一定在当前星星的上边和右边，如此，只需每次输入星星时查询该星星的等级并记录下。

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#define maxn 32000+10#define maxi 15000+10using namespace std;int sum[maxn<<2];int grad[maxi];/*struct lnode{    int x,y;};lnode star[maxi];int cmp(lnode a,lnode b){    if(a.y!=b.y)        return a.y<b.y;    return a.x<b.x;}*/void pushup(int o){    sum[o]=sum[o<<1]+sum[o<<1|1];}void build(int o,int l,int r){    sum[o]=0;    if(l==r)        return ;    int mid=(l+r)>>1;    build(o<<1,l,mid);    build(o<<1|1,mid+1,r);    pushup(o);}void update(int o,int l,int r,int aim){    if(l==r)    {        sum[o]++;        return ;    }    int mid=(l+r)>>1;    if(aim<=mid)        update(o<<1,l,mid,aim);    else        update(o<<1|1,mid+1,r,aim);    pushup(o);}int query(int o,int l,int r,int a,int b){    if(a<=l&&b>=r)        return sum[o];    int mid=(l+r)>>1;    int ans=0;    if(b<=mid)        ans+=query(o<<1,l,mid,a,b);    else if(a>mid)        ans+=query(o<<1|1,mid+1,r,a,b);    else        ans+=query(o<<1,l,mid,a,mid)+query(o<<1|1,mid+1,r,mid+1,b);    return ans;}int main(){    int n,xi,yi;    while(~scanf("%d",&n))    {        build(1,1,maxn);        memset(grad,0,sizeof(grad));       /* for(int i=0;i<n;++i)            scanf("%d%d",&star[i].x,&star[i].y);        sort(star,star+n,cmp);        for(int i=0;i<n;++i)        {            int tem=query(1,1,maxn,1,star[i].x+1);            grad[tem]++;            update(1,1,maxn,star[i].x+1);        }*/        for(int i=0;i<n;++i)        {            scanf("%d%d",&xi,&yi);            int tem=query(1,1,maxn,1,xi+1);            grad[tem]++;            update(1,1,maxn,xi+1);        }        for(int i=0;i<n;++i)            printf("%d\n",grad[i]);    }    return 0;}