poj-2155-Matrix（树状数组 || 线段树）
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
树状数组
二维的写起来很方便，两重循环。
如果是要修改(x1,y1) - (x2,y2)的矩形区域。
那么可以在(x1,y1) 出加1，在(x2+1,y1)处加1，在(x1,y2+1)处加1，在(x2+1,y2+1)处加1
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#define N 1005
using namespace std;
int c[N][N], n;
void add(int x, int y, int d)
{
for (int i = x; i <= n; i += i&-i)
for (int j = y; j <= n; j+=j&-j)
c[i][j] += d;
}
int sum(int x, int y)
{
int ans = 0;
for (int i = x ; i > 0; i-=i&-i)
for (int j = y; j > 0; j-=j&-j)
ans += c[i][j];
return ans;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("1.txt", "r", stdin);
#endif
int T, X, x, xx, y, yy;
char cc;
cin >> X;
while(X--)
{
cin >> n >> T;
memset(c, 0, sizeof(c));
while(T--)
{
cin >> cc;
if (cc == 'C')
{
cin >> x >> y >> xx >> yy;
add(x, y, 1);
add(xx+1, y, 1);
add(x, yy+1, 1);
add(xx+1, yy+1, 1);
}
else
{
cin >> x >> y;
cout << (sum(x, y)&1) << endl;
}
}
if (X) cout << endl;
}
return 0;
}线段树
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cmath>
#include <map>
#include <algorithm>
#define N 1005
#define ll long long
#define mod 9901
const int mm = 1000000007;
int n, num, x, y, xx, yy;;
bool tree[N<<2][N<<2];
using namespace std;
void editY(int kx, int ky, int yl, int yr)
{
if (y <= yl && yy >= yr)
{
tree[kx][ky] ^= 1;
return ;
}
int mid = (yl+yr)>>1;
if (y <= mid) editY(kx, ky<<1, yl, mid);
if (yy > mid) editY(kx, ky<<1|1, mid+1, yr);
}
void editX(int kx, int xl, int xr)
{
if (x <= xl && xx >= xr)
{
editY(kx, 1, 1, n);
return ;
}
int mid = (xl+xr)>>1;
if (x <= mid) editX(kx<<1, xl, mid);
if (xx > mid) editX(kx<<1|1, mid+1, xr);
}
void queryY(int kx, int ky, int yl, int yr)
{
if (tree[kx][ky]) num++;
if (yl==yr) return ;
int m = (yl+yr)>>1;
if (y <= m) queryY(kx, ky<<1, yl, m);
else queryY(kx, ky<<1|1, m+1, yr);
}
void queryX(int kx, int xl, int xr)
{
queryY(kx, 1, 1, n);
if (xl == xr) return ;
int m = (xl+xr)>>1;
if (x <= m) queryX(kx<<1, xl, m);
else queryX(kx<<1|1, m+1, xr);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("1.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
// cin.tie(0);
int i, j, X, T;
char c;
cin >> X;
while(X--)
{
memset(tree, 0, sizeof(tree));
cin >> n >> T;
while(T--)
{
cin >> c;
if (c == 'C')
{
cin >> x >> y >> xx >> yy;
editX(1, 1, n);
}
else
{
cin >> x >> y;
num = 0;
queryX(1, 1, n);
cout << (num&1) << endl;
}
}
if (X) cout << endl;
}
return 0;
}