HDU 1542 STARS [tree array] [data structure]

#include < span class="hljs-keyword">using namespace std;/**************************************/< span class="hljs-keyword">const int N = 50000 + 5;#define lowbit(x) (x&(-x)) int sum[N],cnt,nnn[N];void update(int index ,int val){ for(int i=index;i<=N;i+=lowbit(i)){ sum[i]+=val; }}int getSum(int index) {int ans = 0; for (int i = index; i>0; i -= lowbit( i)) ans += sum[i]; return ans;}int main(){ while(~scanf("%d", &cnt)){ int x,y; memset(sum,0,sizeof(sum)); memset(nnn,0,sizeof(nnn)); for(int i=0;iscanf span>("%d%d",&x,&y); nnn[getSum(x+1)]++; update (x+1,1);} for(int i=0;iprintf("%d
",nnn[i]);}} return span> 0;}

Topic link: http://acm.hdu.edu. cn/showproblem.php?pid=1541
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Stars

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8916 Accepted Submission(s): 3547

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.< /p>

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 o utput 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

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The main idea of ​​the title:
is to find the number of stars in the lower left corner

The idea of ​​solving the problem: < br> is a basic data structure
Both tree arrays and line segment trees can be used.
The tree array AC I use is free to review and it is also rewarding when maintaining i<=maxn instead of The length of the tree array. . It's very strong. I wonder if I didn't study the tree array well, I went to build a line segment tree. . Then I gave up the data structure after doing it for a few days. . .

Read the question carefully and find that y is increasing, so the value of y is useless. We only need to maintain x and record it with a bucket before maintenance;;

Attach this question Code
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#include < span class="hljs-keyword">using namespace std;/**************************************/< span class="hljs-keyword">const int N = 50000 + 5;#define lowbit(x) (x&(-x)) int sum[N],cnt,nnn[N];void update(int index ,int val){ for(int i=index;i<=N;i+=lowbit(i)){ sum[i]+=val; }}int getSum(in t index) {int ans = 0; for (int i = index; i>0; i -= lowbit(i )) ans += sum[i]; return ans;}int main(){ while(~scanf("%d",&cnt )){ int x,y; memset(sum, 0,sizeof(sum)); memset(nnn,0,sizeof(nnn)); for(int i=0;iscanf("%d%d",&x,&y); nnn[getSum(x+1)]++; update (x+1,1);} for(int i=0;iprintf("%d
",nnn[i]);}} return span> 0;}