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 程式師世界 >> 編程語言 >> C語言 >> C++ >> 關於C++ >> POJ 1195 Mobile phones(二維樹狀數組)

POJ 1195 Mobile phones(二維樹狀數組)

編輯:關於C++
Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 15968   Accepted: 7373

Description

Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contains a base station. The number of active mobile phones inside a square can change because a phone is moved from a square to another or a phone is switched on or off. At times, each base station reports the change in the number of active phones to the main base station along with the row and the column of the matrix.

Write a program, which receives these reports and answers queries about the current total number of active mobile phones in any rectangle-shaped area.

Input

The input is read from standard input as integers and the answers to the queries are written to standard output as integers. The input is encoded as follows. Each input comes on a separate line, and consists of one instruction integer and a number of parameter integers according to the following table.
\

The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3.

Table size: 1 * 1 <= S * S <= 1024 * 1024
Cell value V at any time: 0 <= V <= 32767
Update amount: -32768 <= A <= 32767
No of instructions in input: 3 <= U <= 60002
Maximum number of phones in the whole table: M= 2^30

Output

Your program should not answer anything to lines with an instruction other than 2. If the instruction is 2, then your program is expected to answer the query by writing the answer as a single line containing a single integer to standard output.

Sample Input

0 4
1 1 2 3
2 0 0 2 2 
1 1 1 2
1 1 2 -1
2 1 1 2 3 
3

Sample Output

3
4

Source

IOI 2001



題意:第一行輸入兩個數據id,n,分別代表編號和矩陣的大小,後面的數據每一行第一個數據代表編號,1代表會輸入3個數想,有,z代表將矩陣中位置行x,列為y的元素加z(初始化矩陣中所有的元素都為0),2代表會輸入4個數xx,yy,x,y。代表輸出左上角坐標為x,y,右下角坐標為xx,yy的矩陣中所有元素的和


#include
#include
#include
#include
#include

using namespace std;

const int maxn = 1025;

int n,id;
int c[maxn][maxn];

int lowbit(int x)
{
    return x&(-x);
}

void updata(int i,int j,int k)
{
    while(i<=n)
    {
        int pj = j;
        while(pj<=n)
        {
            c[i][pj] += k;
            pj += lowbit(pj);
        }
        i += lowbit(i);
    }
}

int getsum(int i,int j)
{
    int sum = 0;
    while(i>0)
    {
        int pj = j;
        while(pj>0)
        {
            sum += c[i][pj];
            pj -= lowbit(pj);
        }
        i -= lowbit(i);
    }
    return sum;
}

int main()
{
    scanf("%d%d",&id,&n);
    int x,y,z,xx,yy;
    while(scanf("%d",&id)!=EOF)
    {
        if(id == 3)
        {
            break;
        }
        if(id == 1)
        {
            scanf("%d%d%d",&x,&y,&z);
            x++;
            y++;
            updata(x,y,z);
        }
        else if(id == 2)
        {
            scanf("%d%d%d%d",&x,&y,&xx,&yy);
            x++;
            y++;
            xx++;
            yy++;
            printf("%d\n",getsum(xx,yy)-getsum(x-1,yy) - getsum(xx,y-1) + getsum(x-1,y-1));
        }
    }
    return 0;
}



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