POJ 2155 Matrix

题目链接：

解题思路：首先考虑一维的情况，更改某个区间[l, r]时相当于l出加一，r + 1除减一，这样就能保证每次计算的时候两者总是能抵消的。当然也可以在l与r + 1除都加一，而每次求和之后mod 2。

接下来考虑二维，更改区间[x1, y1], [x2, y2]的时候，只需在(x1, y1), (x1, y2 + 1), (x2 + 1, y1), (x2 + 1, y2 + 1)四个点除加一即可（好吧其实我只推出了一维，二维是看别人博客的）。然后求和就交给二维树状数组了。

比较有意思的是，当一个函数这样写的时候

void add(int i, int j, int val){
    for(i; i <= n; i += lowbit(i)){
        for(j; j <= n; j += lowbit(j)){
            bit[i][j] += val;
        }
    } 
}

得到的结果竟然是错误的，不知道是什么bug。

代码：

#define _CRT_SECURE_NO_WARNINGS
#include <functional>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <cctype>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const ll mod = 1000000007;
ll powmod(ll a, ll b) { ll res = 1; a %= mod; assert(b >= 0); for (; b; b >>= 1) { if (b & 1)res = res*a%mod; a = a*a%mod; }return res; }
// head
const int inf = 0x3f3f3f3f;
const int maxn = 1010;

int bit[maxn][maxn], n;

int lowbit(int x){
    return x & (-x);
}
int sum(int x, int y){
    int ans = 0;
    for(int i = x; i > 0; i -= lowbit(i)){
        for(int j = y ; j > 0; j -= lowbit(j)){
            ans += bit[i][j];
        }
    }
    return ans;
}
void add(int x, int y, int val){
    for(int i = x; i <= n; i += lowbit(i)){
        for(int j = y; j <= n; j += lowbit(j)){
            bit[i][j] += val;
        }
    } 
}

int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        memset(bit, 0, sizeof(bit));
        int m;
        scanf("%d %d", &n, &m);
        while(m--){
            char ch;
            scanf(" %c", &ch);
            if(ch == 'C'){
                int x1, y1, x2, y2;
                scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
                
                add(x1, y1, 1); 
                add(x2 + 1, y1, 1);
                add(x1, y2 + 1, 1);
                add(x2 + 1, y2 + 1, 1);
                
            }
            else{
                int x1, y1;
                scanf("%d %d", &x1, &y1);
                if(sum(x1, y1) % 2 == 0) puts("0");
                else puts("1");
            }
        }
        if(T)
            puts("");
    }
}

题目：

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 28786		Accepted: 10494

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

Source

POJ Monthly,Lou Tiancheng