HDU 1541_Stars

题目描述：

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.

题意就是给你n个星星的坐标，并且会按y的升序输入，并把这个星星左下角的星星的数目规定这个星星的等级，最后输出各等级星星的数目；

样例：

input:

51 15 17 13 35 5

output:

12110

树状数组的模版入门题，至于树状数组，你可以简单理解成一个没有各节点右儿子的线段树，具体的解释在挑程那本书上写的已经非常清楚了。

AC代码

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<stack>#include<queue>#include<algorithm>using namespace std;const int MAXM=20010;int n;int bit[4*MAXM];int lev[4*MAXM];void add(int x,int val){    while(x<MAXM*4)    {        bit[x]+=val;        x=x+(x&(-x));    }}int sum(int x){    int t=0;    while(x>0)    {        t+=bit[x];        x=x-(x&(-x));    }    return t;}int main(){    while(scanf("%d",&n)!=EOF)    {        memset(bit,0,sizeof(bit));        memset(lev,0,sizeof(lev));        for (int i=1;i<=n;i++)        {            int x,y;            scanf("%d%d",&x,&y);            lev[sum(++x)]++;            add(x,1);        }        for (int i=0;i<n;i++)            printf("%d\n",lev[i]);    }    return 0;}