ABC-407-D - Domino Covering XOR（dfs优化）

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。

// 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

源代码

// 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……）