poj 2352 && hdu 1541 Stars（树状数组）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1541点击打开链接

Stars

                                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

题意：给出一些星星的横坐标和纵坐标，而且星星的纵坐标按非递减排列，如果纵坐标相等，则横坐标按递增排列，任意两颗星星不会重合。如果有n颗星星的横坐标比某颗星星小而且纵坐标不大于那颗星星（即有n颗星星位于那颗星星的左下角或者左边）则此星星的等级为n，最后输出等级为0至n-1的星星的数量。树状数组 注意每次输入需要将x+1 才表示小于的个数

#include <stdio.h>#include <stdlib.h>#include <iostream>#include<algorithm>#include <math.h>#include <string.h>#include <limits.h>#include <string>#include <queue>#include <stack>using namespace std;int ans[111111];int c[111111];int lowbit(int x){return x&(-x);}void add(int k){    if(k==0)    {        while(k<40000)        {            c[k]++;            k++;        }    }    else    {    while(k<40000)    {        c[k]++;        k+=lowbit(k);    }    }}int read(int k){    int s=0;    if(k==0)        return s;    while(k>0)    {        s+=c[k];        k-=lowbit(k);    }    return s;}int main(){    int n=0;    while(~scanf("%d",&n))    {        memset(c,0,sizeof(c));        memset(ans,0,sizeof(ans));    int x,y;    for(int i=0;i<n;i++)    {        scanf("%d%d",&x,&y);        ans[read(x+1)]++;        add(x+1);    }    for(int i=0;i<n;i++)    {        printf("%d\n",ans[i]);    }    }}