hdu1541-Stars(树状数组)

Stars

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6453    Accepted Submission(s): 2561

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

 需要了解树状数组的特征与构成。

即使知道本题主题是树状数组还是半天想不到，最后根据y的升序输入想到的，题目这样说明也算变相提示了吧。

已知输入数据服从y的升序，且y相同的时候服从x升序。那么可以很明显的想到 对于由输入的到的多组数据x[i],y[i]中，设当前得到的为x[n],y[n]，则y[n]为纵坐标最大值，所有y[i]<=y[n]。

因此只要计算之前数据中x[i]<=x[n]即可。所有x[i]<=x[n]的点都算作x[n]的级数。此处应用树状数组求和。 

而为了保证后效性，此处也应该对父节点修改，用到了树状数组对父节点修改的便利性。

代码：

#include <cstdio>#include <algorithm>#include <string>#include <cstring>#include <stack>#include <cmath>#include <queue>#include <list>#include <cstdlib>#include <vector>#include <set>#include <map>#include <sstream>#include <iostream>#include <stack>using namespace std;int a[32005],c[32005];int lowbit(int x){    return x&(-x);}void add(int x,int num){    while (x<=32000) {        c[x]+=num;        x+=lowbit(x);    }}int sum(int x){    int s=0;    while (x) {        s+=c[x];        x-=lowbit(x);    }    return s;}int main(){    int n,x,y;    while (cin>>n) {        memset(c, 0, sizeof(c));        memset(a, 0, sizeof(a));        for (int i=0; i<n; i++) {            scanf("%d %d",&x,&y);            a[sum(x+1)]++;            add(x+1,1);        }        for (int i=0; i<n; i++) {            printf("%d\n",a[i]);        }    }    return 0;}