POJ 2155 Matrix （二维树状数组）

Matrix

Time Limit:3000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

Submit         Status

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

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

Sample Output

1
0
0
1

这是楼教主出的经典题目啊~~~不A别后悔呦^_^

题意：给个N*N的矩阵，所有元素初始化为零，且该矩阵里的元素只能有0，1。有两个操作——改变和查询，改变是改变一个指定的矩阵（会给出左上角和右下角的坐标）里的

所有元素，即0变为1，1变为0。查询是查询矩阵中某个元素（会给出坐标）的值。

心得：细节没注意，TLE到吐血！！！最开始也想到了方法，但是到后来写出来的时候一直TLE，郁闷。。。后来一看好多小细节！比如说读取操作类型的时候，最好不要直接开一个char变量f然后用%c读入，这样很容易出问题，可以开一个字符数组f[4]用%s读入，判断的时候用f[0]去比较即可。

分析：

         思路一：直接按要求改变元素的值，最后输出即可。不过此时的sum（）要有所改变，不是加了，而是抑或。change（）里面，取反时可以用‘  ！’取反，也可以&1。

详见代码1

  

         思路二：由于初始所有元素均为0，则可用sum（）记录每个元素改变的次数，只需%2判断即可确定每个元素的值。详见代码2

AC代码1:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define INF 0x7fffffff
using namespace std;
const int maxn = 1000 + 10;
int c[maxn][maxn];
int n;

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

void change(int x,int y)
{
    int i,j;
    for(i=x; i<=n; i+=lowbit(i))
        for(j=y; j<=n; j+=lowbit(j))
            c[i][j] = !c[i][j];   
//            c[i][j] ^= 1;            //和上面的取反均可   
}

int sum(int x,int y)
{ 
    int ret = 0,i,j; 
    for(i=x; i>0; i-=lowbit(i))
        for(j=y; j>0; j-=lowbit(j))
            ret ^= c[i][j];           //此时不再是+，而是^ 
    return ret;
}

int main()
{ 
    int T,t,x1,y1,x2,y2;
    char f[4];
    scanf("%d",&T);
    while(T--)
    {
        memset(c,0,sizeof(c));
        scanf("%d%d",&n,&t);
        while(t--)
        {
            scanf("%s",f);
            if(f[0] == 'C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                change(x1,y1);          //此处很容易错，把图画出来就很清晰了
                change(x1,y2+1);
                change(x2+1,y1);
                change(x2+1,y2+1);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                printf("%d\n",sum(x1,y1));
            }
        }
        printf("\n");
    }
    return 0;
}

AC代码2：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define INF 0x7fffffff
using namespace std;
const int maxn = 1000 + 10;
int c[maxn][maxn];
int n;

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

void change(int x,int y)
{
    int i,j;
    for(i=x; i<=n; i+=lowbit(i))
        for(j=y; j<=n; j+=lowbit(j))
            c[i][j]++;                 //只需++，用来计改变的次数
}

int sum(int x,int y)
{
    int ret = 0,i,j;
    for(i=x; i>0; i-=lowbit(i))
        for(j=y; j>0; j-=lowbit(j))
            ret += c[i][j];
    return ret;
}

int main()
{
    int T,t,x1,y1,x2,y2;
    char f[4];
    scanf("%d",&T);
    while(T--)
    {
        memset(c,0,sizeof(c));
        scanf("%d%d",&n,&t);
        while(t--)
        {
            scanf("%s",f);
            if(f[0] == 'C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                change(x1,y1);
                change(x1,y2+1);
                change(x2+1,y1);
                change(x2+1,y2+1);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                printf("%d\n",sum(x1,y1)%2);            //%2判断，确定元素的值
            }
        }
        printf("\n");
    }
    return 0;
}