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POJ 2155 Matrix 二维树状数组

2024-08-06 06:35:52   219人阅读

Matrix
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 19174 Accepted: 7207

Description

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N).

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a ‘0‘ then change it into ‘1‘ otherwise change it into ‘0‘). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
2. Q x y (1 <= x, y <= n) querys A[x, y].

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case.

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.

Output

For each querying output one line, which has an integer representing A[x, y].

There is a blank line between every two continuous test cases.

Sample Input

12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1

Sample Output

1001
Note 
这是一个二维的树状数组，区间更新，单点查询
#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <algorithm>using namespace std;const int MAX = 1010;int c[MAX][MAX];int n;int a,b,e,d;int Lowbit(int x){    return x & (-x);}void Updata(int x,int y,int z){    int i,k;    for(i=x; i<=n; i+=Lowbit(i))        for(k=y; k<=n; k+=Lowbit(k))            c[i][k]+=z;}int Get(int x,int y){    int i,k,sum = 0;    for(i=x; i>0; i-=Lowbit(i))        for(k=y; k>0; k-=Lowbit(k))            sum += c[i][k];    return sum;}int main(){    int x;    scanf("%d",&x);    while(x--)    {        memset(c,0,sizeof(c));        int q;        scanf("%d%d",&n,&q);        char s;        while(q--)        {            //cout<<q<<endl;            scanf(" %c",&s);            if(s==‘C‘)            {                scanf("%d%d%d%d",&a,&b,&e,&d);                a++,b++,e++,d++;                Updata(e,d,1);                Updata(a-1,d,-1);                Updata(e,b-1,-1);                Updata(a-1,b-1,1);            }            if(s==‘Q‘)            {                scanf("%d%d",&a,&b);                printf("%d\n",Get(a,b)%2);            }        }        if(x!=0)            printf("\n");    }    return 0;}

 

POJ 2155 Matrix 二维树状数组


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