[Codeforces Round #498 (Div. 3)] -F. Xor-Paths （折半搜索）

[Codeforces Round #498 (Div. 3)] -F. Xor-Paths （折半搜索）

F. Xor-Paths

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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：

给定一个\(n*m\)的矩阵，第\(\mathit i\) 行 第\(\mathit j\)列的数值为\(a[i][j]\)，现在让你算出有多少个从\((1,1)\)到\((n,m)\)只向下或者向右的路径中经过的数值异或和为\(\mathit k\)，注意：路径不能走出矩阵。

思路：

观察数据范围\(n\leq 20,m\leq 20\)，显然直接dfs搜索的时间复杂度为\(O(2^{n+m})\)肯定会超时的。

因为$x\oplus y= k \rightarrow x=y\oplus k $，

我们可以将\((1,1)\)向右下方向走到\(x+y=n+1\)的那条斜线上（当\(n=m\)时是对角线），

并用\(map<ll,ll> a[][]\) 记录\(a[x][y][z]\)代表走到\((x,y)\)位置时异或值为\(\mathit z\)的路径个数。

将\((n,m)\)向左上方向也走到\(x+y=n+1\)的那条斜线上，如果\((n,m)\)走来的异或值为\(x\),

那么该路径对答案的贡献为\(a[x][y][x\oplus k]\)，将所有路径的贡献做个和就是答案。

时间复杂度分析：

在右下方向是从\((x,y)\) 最多只会走向\((x+1,y),(x,y+1)\)，显然\(x+y\)的值都增一，

当\((x+y)=n+1\)时dfs停止，所以这部分的时间复杂度为\(O(2^{n} )\)

同理得 从\((n,m)\)开始的路径部分的时间复杂度为\(O(2^{m-2})\)

所以总的时间复杂度为\(O((2^{n}+2^{m-2})*log_2(2^{n}+2^{m-2}))\)

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#include <sstream>
#include <bitset>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
map<ll, ll> b[22][22];
ll a[22][22];
int n, m;
ll k;
ll ans = 0ll;
void dfs1(int x, int y, ll now)
{
    if (x + y == n + 1)
    {
        now ^= a[x][y];
        b[x][y][now] += 1;
        return ;
    }
    if (x < n)
        dfs1(x + 1, y, now ^ a[x + 1][y]);
    if (y < m)
        dfs1(x, y + 1, now ^ a[x][y + 1]);
}
void dfs2(int x, int y, ll now)
{
    if (x + y == n + 1)
    {
        ans += b[x][y][k ^ now];
        return ;
    }
    if (x > 1)
        dfs2(x - 1, y, now ^ a[x - 1][y]);
    if (y > 1)
        dfs2(x, y - 1, now ^ a[x][y - 1]);
}

int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    n = readint();
    m = readint();
    k = readll();
    repd(i, 1, n)
    {
        repd(j, 1, m)
        {
            a[i][j] = readll();
        }
    }
    dfs1(1, 1, a[1][1]);
    dfs2(n, m, a[n][m]);
    printf("%lld\n", ans );
    return 0;
}