HDU 2642 Stars [Two-dimensional tree array] [data structure]

#include #define abs(x) (((x)>0)?(x): -(x))#define lalal puts("*********") #define Rep(a,b,c) for(int a=(b);a<=(c);a++)#define Req(a,b, c) for(int a=(b);a>=(c);a--)#define Rop(a,b,c) for(int a =(b);a<(c);a++)#define s1(a) scanf("%d",&a)typedef long long int LL;using namespace std< /span>;/*********************************** ***/const int N =  1000+5;#define lowbit(x) (x&-x)LL sum[N][N];bool h[N][N];void update(int xi,int yi,int val){ for (int i=xi;i<=N;i+=lowbit(i)) for(int j=yi;j<=N;j+=lowbit(j)) sum[i][j]+=val; return;}int getSum(int xi,int yi){ int ans = 0; for(int i=xi;i>0;i -=lowbit(i)) for(int j=yi;j>0;j-=lowbit(j)) ans+=sum[i][ j]; return ans ;}int main(){  int n; while(~s1(n)){ memset(sum,0,sizeof(sum)); memset( h,0,sizeof(h)); char< /span> ch; int x1,y1,x2,y2; int tem; while(n--){ getchar(); scanf("%c" ,&ch); if('B'==ch){ s1(x1),s1(y1); x1++,y1++; if(h[x1][y1]) continue< /span>; update(x1,y1,1); h[x1][y1]=1 ;} else if('D' ==ch){ s1(x1),s1(y1); x1++,y1++; if(h[x1][y1]) update(x1,y1,- 1); h[x1][y1]=0;} else {s1(x1),s1(x2),s1(y1),s1(y2); x1++,y1++,x2++,y2++; if(x1>x2) swap(x1,x2); if(y1>y2) swap(y1,y2); tem=getSum(x2,y2)-getSum (x2,y1-1)-getSum(x1-1,y2)+getSum(x1-1,y1-1); printf("%d
",tem);}}} return 0 ;}

Subject link: http://acm.hdu.edu.cn/showproblem.php?pid=2642

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Stars

Time Limit: 5000/2000 MS (Java /Others) Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 1675 Accepted Submission(s): 706

Problem Description
Yifenfei is a romantic guy and he likes to count the stars in the sky.
To make the problem easier, we considerate the sky is a two-dimension plane.Sometimes the star will be bright and sometimes the star will be dim.At first,there is no bright star in the sky, then some information will be given as “B xy” where’B’ rep resent bright and x represent the X coordinate and y represent the Y coordinate means the star at (x,y) is bright,And the’D’ in “D xy” mean the star at(x,y) is dim.When get a query as “Q X1 X2 Y1 Y2”,you should tell Yifenfei how many bright stars there are in the region correspond X1,X2,Y1,Y2.

There is only one case.

Input
The first line contain a M(M <= 100000), then M line followed.
each line start with a operational character.
if the character is B or D, then two integer X,Y (0 <=X,Y<= 1000) followed.
if the character is Q then four integer X1,X2,Y1,Y2(0 <=X1,X2,Y1,Y2<= 1000) followed .

Output
For each query, output the number of bright stars in one line.

Sample Input
5
B 581 145
B 581 145
Q 0 600 0 200
D 581 145
Q 0 600 0 200

Sample Output
1
0

Author
teddy

Source
Millions of Qin Guan will eventually belong to Chu

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The main idea of ​​the topic:
There are a total of M operations
There are a total of 3 operations
1) B xy A star appears at the position (x, y)
2) The star at the position D xy (x, y) disappears
3) Q x1 x2 y1 y1 Query how many stars are in the rectangle

< p>Question-solving idea:

There is nothing to say in the idea, because it is dynamic, it must be a two-dimensional tree array, which can be maintained.

Note that the same coordinates There can only be one star in the bottom. If you delete it, there will be no stars in this coordinate.

Note When doing the question, pay attention to the range of the coordinates, because it starts from 0 when calculating All coordinates must be added by 1, otherwise there will be infinite TLE

Attach the code of this question
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#include #define abs(x) (((x)>0)?(x):-(x))#define lalal puts("***** ****")#define Rep(a,b,c) for(int a=(b);a<=(c);a++)#define Req(a,b,c) for(int a=(b);a>=(c);a--)#define Rop(a,b,c) for(int a=(b);a<(c);a++)#define s1 (a) scanf("%d",&a)typedef long long i nt LL;using namespace std span>;/************************************ **/const int N = 1000 +5;#define lowbit(x) (x&-x)LL sum[N ][N];bool h[N][N];void update(int xi,int yi,int val){ for(int i=xi;i<=N;i+=lowbit(i)) for(int j=yi;j<=N;j+=lowbit(j)) sum[i][j]+= val; return;}int getSum(int xi,int yi){ int ans = 0; for(int i=xi;i> 0;i-=lowbit(i)) for( int j=yi;j>0;j-=lowbit(j)) ans+=sum[i][j]; return ans ;}int main(){ int n; < span class="hljs-keyword">while(~s1(n)){ memset(sum, 0,sizeof(sum)); memset(h,0,sizeof(h)); char ch; i nt x1,y1,x2,y2; int tem; while(n--) {getchar(); scanf("%c",&ch); if('B'==ch){ s1(x1),s1(y1); x1++,y1++; if(h[x1][y1]) continue; update(x1,y1,1); h[x1][y1]=1;} else if('D'==ch){ s1(x1),s1(y1); x1++,y1++; if(h[x1][y1]) update(x1,y1,-1); h[ x1][y1]=0; } else {s1(x1),s1(x2),s1(y1),s1(y2); x1++,y1++,x2++,y2++; if(x1>x2) swap(x1,x2); if(y1>y2) swap(y1,y2); tem =getSum(x2,y2)-getSum(x2,y1-1)-getSum(x1-1 ,y2)+getSum(x1-1,y1-1); printf("%d
",tem);}}} return 0;}