Codeforces Round #498 (Div. 3) F. Xor-Paths

题目链接：F. Xor-Paths

题解：从起点和终点双向搜索在中间相遇时更新答案

 1 #include<bits/stdc++.h>
 2 #include<set>
 3 #include<cstdio>
 4 #include<iomanip>
 5 #include<iostream>
 6 #include<string>
 7 #include<cstring>
 8 #include<algorithm>
 9 #define pb push_back
10 #define ll long long
11 #define fi first
12 #define se second
13 #define PI 3.14159265
14 #define ls l,m,rt<<1
15 #define rs m+1,r,rt<<1|1
16 #define eps 1e-7
17 #define pii pair<int,int>
18 typedef unsigned long long ull;
19 const int mod=998244353;
20 const ll inf=0x3f3f3f3f3f3f3f;
21 const int maxn=25;
22 using namespace std;
23 ll n,m,cn,ans,k;
24 ll a[25][25];
25 map<ll,ll>mp[maxn][maxn];
26 void dfs1(int x,int y,ll s)
27 {
28     //cout<<x<<" "<<y<<" "<<s<<endl;
29     if(x+y-2==(n+m-2)/2)
30     {
31         //cout<<x<<" "<<y<<" "<<s<<endl;
32         mp[x][y][s]++;return ;
33     }
34     if(x+1<=n)dfs1(x+1,y,s^a[x+1][y]);
35     if(y+1<=m)dfs1(x,y+1,s^a[x][y+1]);
36 }
37 void dfs2(int x,int y,ll s)
38 {
39     if(mp[x][y].size()>0)
40     {
41         //cout<<x<<" "<<y<<" "<<(s^a[x][y])<<" "<<(s^a[x][y]^k)<<endl;
42         ans+=mp[x][y][s^k^a[x][y]];
43         return ;
44     }
45     if(x-1>=1)dfs2(x-1,y,s^a[x-1][y]);
46     if(y-1>=1)dfs2(x,y-1,s^a[x][y-1]);
47 }
48 int main()
49 {
50     ios::sync_with_stdio(false);
51     cin.tie(0);cout.tie(0);
52     cin>>n>>m>>k;
53     for(int i=1;i<=n;i++)
54     {
55         for(int j=1;j<=m;j++)
56         cin>>a[i][j];
57     }
58     dfs1(1,1,a[1][1]);
59     dfs2(n,m,a[n][m]);
60     cout<<ans<<endl;
61     return 0;
62 }

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