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
3 3 112 1 57 10 012 6 4
Output
3
Input
3 4 21 3 3 30 3 3 23 0 1 1
Output
5
Input
3 4 10000000000000000001 3 3 30 3 3 23 0 1 1
Output
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) .
题目：G - Xor-Paths
思路：
折半思想，前一半：从位置(1,1)开始到x+y=(n+m)/2,可以看成函数，进行异或。
后一半：从（n,m)开始，一直到x+y=(n+m)/2,异或。
异或有交换律，还有其他运算法则。
由运算法则可以推出公式
k=a1^a2^...^an;
令i=(1+n)/2;
q=a1^a2^...^ai;
sum=ai^...^an;
k=sum^ai^q;
所以q=sum^ai^k;
然后你第一个函数在走到的点给sum值打个标记，然后第二个函数把sum^ai就是取消这一步的值（这个值在前一个函数）
再^k就是找到跟现在这个sum异或起来为k（也就是合成一条路径）标记的值，也就是方案有多少
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn=25;
ll a[maxn][maxn],ans,k;
map<ll,ll>mp[maxn];
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
int n,m;
void dfs_pre(int x,int y,ll sum)
{
if(x+y==(n+m+2)/2) {mp[x][sum]++;return ;}//x+y==(n+m+2)/2,有个+2是因为有两种特殊情况，一个是n=1，一个是m=1
for(int i=0;i<2;i++)
{
int tx=x+dx[i];
int ty=y+dy[i];
if(tx<1||ty<1||tx>n||ty>m) continue;
dfs_pre(tx,ty,sum^a[tx][ty]);
}
}
void dfs_end(int x,int y,ll sum)
{
if(x+y==(n+m+2)/2) {ans+=mp[x][sum^k^a[x][y]];return ;}
for(int i=2;i<4;i++)
{
int tx=x+dx[i];
int ty=y+dy[i];
if(tx<1||ty<1||tx>n||ty>m) continue;
dfs_end(tx,ty,sum^a[tx][ty]);
}
}
int main()
{
scanf("%d%d%I64d",&n,&m,&k);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
scanf("%I64d",&a[i][j]);
}
}
dfs_pre(1,1,a[1][1]);
dfs_end(n,m,a[n][m]);
printf("%I64d\n",ans);
return 0;
}
原文地址：https://www.cnblogs.com/EchoZQN/p/10351453.html