ABC-407-D - Domino Covering XOR（dfs优化） · znzryb
## 思路讲解
其实还是比较简单的，就是如果说最后的答案不涉及这个顺序问题，其实是可以不用管这个顺序的。
当然，所有选择的可能都是要管的。
```cpp
// 选择1：不放多米诺骨牌，跳过当前格子
vis[i][j] = true;
dfs(covered_xor);
vis[i][j] = false;
// 选择2：尝试垂直放置多米诺骨牌 (向下)
if (i + 1 <= N && !vis[i + 1][j]) {
vis[i][j] = vis[i + 1][j] = true;
dfs(covered_xor ^ A[i][j] ^ A[i + 1][j]);
vis[i][j] = vis[i + 1][j] = false;
}
// 选择3：尝试水平放置多米诺骨牌 (向右)
if (j + 1 <= M && !vis[i][j + 1]) {
vis[i][j] = vis[i][j + 1] = true;
dfs(covered_xor ^ A[i][j] ^ A[i][j + 1]);
vis[i][j] = vis[i][j + 1] = false;
}
```
## AC代码
[https://atcoder.jp/contests/abc407/submissions/66113167](https://atcoder.jp/contests/abc407/submissions/66113167)
::: 源代码
```cpp
// Problem: D - Domino Covering XOR
// Contest: AtCoder - AtCoder Beginner Contest 407AtCoder 初学者竞赛 407
// URL: https://atcoder.jp/contests/abc407/tasks/abc407_d
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
do { \
std::cerr << #variable << ":"; \
for (const auto& elem : variable) { \
std::cerr << elem << " "; \
} \
std::cerr << "\n"; \
} while (0)
#define uniVec(var) \
do { \
sort(var.begin(),var.end());\
var.resize(unique(var.begin(),var.end())-var.begin());\
} while (0)
#define debugN(var,N) \
do{ \
std::cerr<<#var<<":"; \
FOR(i,1,N){ \
std::cerr<<var[i]<<" "; \
} \
std::cerr<<"\n"; \
}while(0)
#define debugMap(variable) \
do { \
std::cerr << #variable << ":\n"; \
for (const auto& pair : variable) { \
std::cerr << " " << pair.first << " => " << pair.second << "\n"; \
} \
std::cerr << std::endl; \
} while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define vecMxx(vec,x) (vec.end()-lower_bound(all(vec),x)) // vec中大于x的数有几个
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr3;
typedef array<ll,2> arr2;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;
ll N,M,Q,X,K,lT,A[25][25];
bool vis[25][25];
ll ans, total_xor;
inline void solve(){
cin >> N >> M;
total_xor = 0;
ans = 0;
CLR(vis, false);
FOR(i, 1, N) {
FOR(j, 1, M) {
cin >> A[i][j];
total_xor ^= A[i][j];
}
}
function<void(ll)> dfs = [&](ll covered_xor) {
// 找到第一个未被覆盖的格子
ll next_i = -1, next_j = -1;
FOR(i, 1, N) {
FOR(j, 1, M) {
if (!vis[i][j]) {
next_i = i;
next_j = j;
goto found;
}
}
}
found:
// 如果所有格子都处理完了，计算得分
if (next_i == -1) {
ll score = total_xor ^ covered_xor;
ans = max(ans, score);
return;
}
ll i = next_i, j = next_j;
// 选择1：不放多米诺骨牌，跳过当前格子
vis[i][j] = true;
dfs(covered_xor);
vis[i][j] = false;
// 选择2：尝试垂直放置多米诺骨牌 (向下)
if (i + 1 <= N && !vis[i + 1][j]) {
vis[i][j] = vis[i + 1][j] = true;
dfs(covered_xor ^ A[i][j] ^ A[i + 1][j]);
vis[i][j] = vis[i + 1][j] = false;
}
// 选择3：尝试水平放置多米诺骨牌 (向右)
if (j + 1 <= M && !vis[i][j + 1]) {
vis[i][j] = vis[i][j + 1] = true;
dfs(covered_xor ^ A[i][j] ^ A[i][j + 1]);
vis[i][j] = vis[i][j + 1] = false;
}
};
dfs(0);
cout << ans << "\n";
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
// cin>>lT;
// while(lT--){
// solve();
// }
solve();
return 0;
}
/*
AC
https://atcoder.jp/contests/abc407/submissions/66113167
*/
```
:::
## 心路历程（WA，TLE，MLE……）