HDU 2642 Stars 二维树状数组

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#hdu #树状数组

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本文介绍了一个关于星空的问题，通过使用二维树状数组解决星星的点亮、熄灭及查询区间内亮星数量的问题。文章提供了完整的AC代码，并详细解释了代码实现细节。

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Stars

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/65536 K (Java/Others)

Total Submission(s): 105 Accepted Submission(s): 49

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 x y" where 'B' represent 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 x y" 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

题目大意：

B 为点亮 x y 这个星星，一开始星星是暗的， D 是熄灭这个星星，Q 是查询这个区间内多少个亮着的星星。

注意 q 后面给出的数据 x1 x2 y1 y2

思路：

裸二维树状数组。

AC代码：

#include<bits/stdc++.h>
using namespace std;
#define N 1005
int C[N][N];
bool falg[N][N];
int sum(int x,int y)
{
    int ret=0,i,j;
    for(i=x; i>0; i-=(i&-i))
        for(j=y; j>0; j-=(j&-j))
            ret+=C[i][j];
    return ret;
}
void add(int x,int y,int d)
{
    int i,j;
    for(i=x; i<N; i+=(i&-i))
        for(j=y; j<N; j+=(j&-j))
            C[i][j]+=d;
}
int main()
{
    int t,n,a,b,c,d,e,i,j,k,v;
    char s[5];
    while(~scanf("%d",&n))
    {
        for(i=1; i<N; ++i)
            for(j=1; j<N; ++j)
            {
                C[i][j]=0;
                falg[i][j]=0;
            }
        while(n--)
        {
            scanf("%s",s);
            if(s[0]=='Q')
            {
                scanf("%d%d%d%d",&a,&c,&b,&d);
                if(a<c) swap(a,c);
                if(b<d) swap(b,d);
                printf("%d\n",sum(a+1,b+1)-sum(a+1,d)-sum(c,b+1)+sum(c,d));
            }
            else if(s[0]=='B')
            {
                scanf("%d%d",&a,&b);
                if(falg[a+1][b+1])
                    continue;
                add(a+1,b+1,1);
                falg[a+1][b+1]=1;
            }
            else if(s[0]=='D')
            {
                scanf("%d%d",&a,&b);
                if(falg[a+1][b+1]==0)
                    continue;
                add(a+1,b+1,-1);
                falg[a+1][b+1]=0;
            }
        }
    }
}