Codeforces 1006F Xor-Paths

题目链接：http://codeforces.com/contest/1006/problem/F

题面：

F. Xor-Paths

time limit per test3 seconds

memory limit per test 256 megabytes

There is a rectangular grid of size n×mn×m. Each cell has a number written on it; the number on the cell (i,ji,j) is ai,jai,j. Your task is to calculate the number of paths from the upper-left cell (1,11,1) to the bottom-right cell (n,mn,m) meeting the following constraints:

• You can move to the right or to the bottom only. Formally, from the cell (i,ji,j) you may move to the cell (i,j+1i,j+1) or to the cell (i+1,ji+1,j). The target cell can't be outside of the grid.
• The xor of all the numbers on the path from the cell (1,11,1) to the cell (n,mn,m) must be equal to kk (xor operation is the bitwise exclusive OR, it is represented as '^' in Java or C++ and "xor" in Pascal).

Find the number of such paths in the given grid.

Input

The first line of the input contains three integers nn, mm and kk (1≤n,m≤201≤n,m≤20, 0≤k≤10180≤k≤1018) — the height and the width of the grid, and the number kk.

The next nn lines contain mm integers each, the jj-th element in the ii-th line is ai,jai,j (0≤ai,j≤10180≤ai,j≤1018).

Output

Print one integer — the number of paths from (1,11,1) to (n,mn,m) with xor sum equal to kk.

Examples

input

Copy

3 3 11
2 1 5
7 10 0
12 6 4

output

Copy

3

input

Copy

3 4 2
1 3 3 3
0 3 3 2
3 0 1 1

output

Copy

5

input

Copy

3 4 1000000000000000000
1 3 3 3
0 3 3 2
3 0 1 1

output

Copy

0

Note

All the paths from the first example:

• (1,1)→(2,1)→(3,1)→(3,2)→(3,3)(1,1)→(2,1)→(3,1)→(3,2)→(3,3);
• (1,1)→(2,1)→(2,2)→(2,3)→(3,3)(1,1)→(2,1)→(2,2)→(2,3)→(3,3);
• (1,1)→(1,2)→(2,2)→(3,2)→(3,3)(1,1)→(1,2)→(2,2)→(3,2)→(3,3).

All the paths from the second example:

• (1,1)→(2,1)→(3,1)→(3,2)→(3,3)→(3,4)(1,1)→(2,1)→(3,1)→(3,2)→(3,3)→(3,4);
• (1,1)→(2,1)→(2,2)→(3,2)→(3,3)→(3,4)(1,1)→(2,1)→(2,2)→(3,2)→(3,3)→(3,4);
• (1,1)→(2,1)→(2,2)→(2,3)→(2,4)→(3,4)(1,1)→(2,1)→(2,2)→(2,3)→(2,4)→(3,4);
• (1,1)→(1,2)→(2,2)→(2,3)→(3,3)→(3,4)(1,1)→(1,2)→(2,2)→(2,3)→(3,3)→(3,4);
• (1,1)→(1,2)→(1,3)→(2,3)→(3,3)→(3,4)(1,1)→(1,2)→(1,3)→(2,3)→(3,3)→(3,4).

题意：

给出一个N * M的图(N,M <= 20)，和K， 问从(1,1)走到(N,M)且路径上的xor值为K的条数有多少条。

题解：

折半枚举， 先从(1, 1)开始限定深搜层数d， 对当前每个(i, j)记录了到达(i,j)且权值为x的路径有多少条。

再从(n, m)搜， 走到一个已经搜过的(i, j)直接算对答案的贡献。不难证明复杂度为O(2 ^ d * log(2^d) + 2 ^ p * log(2^p)),(p+d=n+m,), 当d取20的时候最优。

代码：

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
bool vis[25][25];
ll a[25][25];
unordered_map<ll, ll> c[25 * 25];
int n, m;
ll k, res;
int hxh(int x, int y)
{
    return x * 21 + y;
}


void dfs_pre(int x, int y, ll cur, int deep)
{
    if(x < 1 || y < 1 || x > n || y > m || deep > 20) return ;
    vis[x][y] = 1;
    cur ^= a[x][y];
    c[hxh(x, y)][cur] ++;
    dfs_pre(x + 1, y, cur, deep + 1);
    dfs_pre(x, y + 1, cur, deep + 1);
}

void dfs_suf(int x, int y, ll cur)
{
    if(x < 1 || y < 1 || x > n || y > m) return ;
    if(vis[x][y]) {
        res += c[hxh(x, y)][k ^ cur];
        return ;
    }
    cur ^= a[x][y];
    dfs_suf(x - 1, y, cur);
    dfs_suf(x, y - 1, cur);
}
int main()
{
    cin >> n >> m >> k;
    for(int i = 1; i <= n; i ++) {
        for(int j = 1; j <= m; j ++) {
            cin >> a[i][j];
        }
    }
    dfs_pre(1, 1, 0, 1);
    dfs_suf(n, m, 0);
    cout << res << '\n';
    return 0;
}