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

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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 1005using 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;}

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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 9901const 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;}