SPOJ-1029 MATSUM

MATSUM - Matrix Summation

no tags 

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:14SET 0 0 1SUM 0 0 3 3SET 2 2 12SUM 2 2 2 2SUM 2 2 3 3SUM 0 0 2 2ENDOutput:1121213

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define maxn 1033int c[maxn][maxn], a[maxn][maxn];inline int lowbit(int x){    return x & (-x);}void add(int x, int y, int v){    while(x < maxn){        int cur = y;        while(cur < maxn){            c[x][cur] += v;            cur += lowbit(cur);        }        x += lowbit(x);    }}int query(int x, int y){    int ans = 0;    while(x){        int cur = y;        while(cur){            ans += c[x][cur];            cur -= lowbit(cur);        }        x -= lowbit(x);    }    return ans;}int main(){    int T;    scanf("%d", &T);    while(T--){        int n, x1, y1, x2, y2, v;        scanf("%d", &n);        memset(c, 0, sizeof(c));        memset(a, 0, sizeof(a));        char s[10];        while(scanf("%s", s) != EOF){            if(s[0] == 'E'){                break;            }            if(s[1] == 'E'){                scanf("%d %d %d", &x1, &y1, &v);                x1++; y1++;                add(x1, y1, -a[x1][y1]);                a[x1][y1] = v;                add(x1, y1, v);            }            else{                scanf("%d %d %d %d", &x1, &y1, &x2, &y2);                x1++; y1++; x2++; y2++;                printf("%d\n", query(x2, y2) - query(x1 - 1, y2) - query(x2, y1 - 1) + query(x1 - 1, y1 - 1));            }        }    }}  /*题意：1024*1024的矩阵，需要对单点修改值，查询子矩阵中的数字和。多cases。思路：二维树状数组的裸题。*/