统计矩阵和 二维树状数组 SPOJ 1029 Matrix Summation

http://www.spoj.com/problems/MATSUM/

MATSUM - Matrix Summation

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A N × N matrix is filled with numbers. BuggyD is analyzing the matrix, and he wants the sum of certain submatrices every now and then, so he wants a system where he can get his results from a query. Also, the matrix is dynamic, and the value of any cell can be changed with a command in such a system.

Assume that initially, all the cells of the matrix are filled with 0. Design such a system for BuggyD. Read the input format for further details.

Input

The first line of the input contains an integer t, the number of test cases. t test cases follow.

The first line of each test case contains a single integer N (1 <= N <= 1024), denoting the size of the matrix.

A list of commands follows, which will be in one of the following three formats (quotes are for clarity):

1. "SET x y num" - Set the value at cell (x, y) to num (0 <= x, y < N).
2. "SUM x1 y1 x2 y2" - Find and print the sum of the values in the rectangle from (x1, y1) to (x2, y2), inclusive. You may assume that x1 <= x2 and y1 <= y2, and that the result will fit in a signed 32-bit integer.
3. "END" - Indicates the end of the test case.

Output

For each test case, output one line for the answer to each "SUM" command. Print a blank line after each test case.

Example

Input:
1
4
SET 0 0 1
SUM 0 0 3 3
SET 2 2 12
SUM 2 2 2 2
SUM 2 2 3 3
SUM 0 0 2 2
END

Output:
1
12
12
13

Warning: large Input/Output data, be careful with certain languages

 Submit solution!

思路: 二维树状数组

分析:

1 简单的二维树状数组，注意因为数据量很大TLE了多次，之后把memset改成for循环A了

代码:

#include
#include
#include
#include
using namespace std;

const int MAXN = 1100;

int n;
int mat[MAXN][MAXN];
int treeNum[MAXN][MAXN];

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

void add(int x , int y , int val){
    for(int i = x ; i < MAXN ; i += lowbit(i))
        for(int j = y ; j < MAXN ; j += lowbit(j))
            treeNum[i][j] += val;
}

long long getSum(int x , int y){
    long long ans = 0;
    for(int i = x ; i > 0 ; i -= lowbit(i))
        for(int j = y ; j > 0 ; j -= lowbit(j))
            ans += treeNum[i][j];
    return ans;
}

void solve(){
    char str[10];
    int x , y , val;
    int x1 , y1 , x2 , y2;
    while(scanf("%s" , str) && str[0] != 'E'){
        if(str[1] == 'E'){
            scanf("%d%d%d%*c" , &x , &y , &val); 
            add(x+1 , y+1 , val-mat[x+1][y+1]);
            mat[x+1][y+1] = val;
        }
        else{
            scanf("%d%d%d%d%*c" , &x1 , &y1 , &x2 , &y2); 
            x1++ , y1++ , x2++ , y2++;
            long long ans = getSum(x2 , y2);  
            ans -= getSum(x1-1 , y2);  
            ans -= getSum(x2 , y1-1);  
            ans += getSum(x1-1 , y1-1);  
            printf("%lld\n" , ans);
        }
    }
}

int main(){
    int cas;
    scanf("%d" , &cas);
    while(cas--){
        scanf("%d%*c" , &n); 
        for(int i = 0 ; i <= n ; i++)
            for(int j = 0 ; j <= n ; j++)
                treeNum[i][j] = mat[i][j] = 0;
        solve();
    }
    return 0;
}