HDU 1541 & POJ 2352 Stars (树状数组)
2017-07-25 21:51
519 查看
Stars
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9874 Accepted Submission(s): 3959
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
5
1 1
5 1
7 1
3 3
5 5
Sample Output
1
2
1
1
0
Source
Ural Collegiate Programming Contest 1999
题意:
给你一个二维的星星图,让你求出每个星星的左下有几颗星星并输出有(0-n-1)颗星星的数量。
POINT:
由于题目是按照y递增的顺序给出星星,就直接记录x位置星星有几颗,树状数组记录区间(1-x)星星有几颗就行了。
#include <stdio.h> #include <string.h> #include <iostream> #include <map> #include <algorithm> using namespace std; #define rt x<<1|1 #define lt x<<1 const int N = 32000*10; #define LL long long int sum ; int n; int maxn; int lowbit(int x) { return x&-x; } void add(int x,int c) { while(x<=32001) { sum[x]+=c; x+=lowbit(x); } } int query(int x) { int ans=0; while(x>=1) { ans+=sum[x]; x-=lowbit(x); } return ans; } int main() { int ans[15010]; while(~scanf("%d",&n)) { memset(sum,0,sizeof sum); memset(ans,0,sizeof ans); int x; int y; for(int i=1;i<=n;i++) { scanf("%d %d",&x,&y); ans[query(x+1)]++; add(x+1,1); } for(int i=0;i<n;i++) { printf("%d\n",ans[i]); } } return 0; }
相关文章推荐
- poj-2352 && HDU-1541 --Stars(树状数组)
- POJ 2352 && HDU 1541 Stars (树状数组)
- poj 2352 && hdu 1541 Stars (树状数组)
- poj 2352 && hdu 1541 Stars (树状数组水题)
- poj 2352 && hdu 1541 Stars (树状数组)
- poj 2352 && hdu 1541 Stars(树状数组)
- POJ 2352 && HDU 1541 Stars (树状数组)
- hdu 1541/poj 2352:Stars(树状数组,经典题)
- 树状数组 POJ 2352 HDU 1541 Stars
- poj2352 && hdu1541 Stars(树状数组)
- POJ 2352 Stars + HDU 1556 Color the ball(树状数组单点更新及伪区间更新)
- poj 2352 && hdu 1541 Stars 线段树
- POJ 2481 Cows && POJ 2352 Stars(树状数组妙用)
- POJ 2481 Cows && POJ 2352 Stars(树状数组妙用)
- POJ 2352 && HDU 1541 Stars(BIT)
- poj 2352 OR hdu 1541 Stars(数据结构:树状数组)
- POJ 2352 Stars (线段树&&树状数组)
- POJ 2352 Stars(树状数组)
- poj 2352 stars (树状数组)
- poj2352 Stars(树状数组)