hdu 5465 Clarke and puzzle

题意：

中文题http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=631&pid=1002

解：

 Nim博弈，不过是2维的，并且有更新操作。这里可以用二位树状数组维护。跟POJ1195很像。

#include <iostream>
#include <algorithm>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>

#define CL(arr, val)    memset((arr), (val), sizeof(arr))
#define REP(i, n)       for(i = 0; i < n; ++i)
#define FOR(i, l, h)    for(i = l; i <= h; ++i)
#define FORD(i, h, l)   for(i = h; i >= l; --i)
#define L(x)            x << 1
#define R(x)            x << 1 | 1
#define MID(l, r)       (l + (r - l) >> 1)

using namespace std;


typedef unsigned int uint;
typedef long long LL;
typedef unsigned long long uLL;

const int N = 510;
const double PI = acos(-1.0);
const double eps = 1e-6;
const double INF = 1e9+7;
const int MOD = 1000000007;

int mat[N][N];
int a[N][N];

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

void add(int x, int y, int d) {
    int i, j;
    for(i = x; i < N; i += lowbit(i))
        for(j = y; j < N; j += lowbit(j))
            mat[i][j] ^= d;
}

LL sum(int x, int y) {
    LL res = 0;
    int i, j;
    for(i = x; i > 0; i -= lowbit(i))
        for(j = y; j > 0; j -= lowbit(j))
            res ^= mat[i][j];
    return res;
}

int main() {
    //freopen("data.in", "r", stdin);
    int T;
    int n, m, q;
    int i, j, op, ans;
    int x, y, x_1, y_1, z;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d%d", &n, &m, &q);
        CL(mat, 0);
        FOR(i, 1, n){
            FOR(j, 1, m){
                scanf("%d", &a[i][j]);
                add(i, j, a[i][j]);
            }
        }
        while(q--) {
            scanf("%d", &op);
            switch(op) {
            case 1:
                scanf("%d%d%d%d", &x, &y, &x_1, &y_1);
                x--; y--;
                ans = sum(x_1, y_1) ^ sum(x, y_1) ^ sum(x_1, y) ^ sum(x, y);    //这里需要注意
                if(ans != 0)   puts("Yes");
                else    puts("No");
                break;
            case 2:
                scanf("%d%d%d", &x, &y, &z);
                if(z == a[x][y])    continue;
                add(x, y, a[x][y]);
                add(x, y, z);
                a[x][y] = z;
                break;
            default:
                break;
            }
        }
    }
    return 0;
}

 

i,j​​。