[CF1006F] Xor-Paths

[CF1006F] Xor-Paths

Description

给出一个 n × m 的网格，每个格子上有权值 (a[i][j])，要从 (1,1) 走到 (n,m)，每次只能向右或向下走，沿路计算异或和，求异或和等于 k 的路径数。

Solution

双向 BFS，由于要走的总步数为 (steps=n+m-2)，正着走 (steps/2)，反着走 ((steps+1)/2)，然后合起来用 map 算答案即可

#include <bits/stdc++.h>
using namespace std;

#define int long long

const int N = 105;

int a[N][N], n, m, k;

signed main()
{
    ios::sync_with_stdio(false);

    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> a[i][j];
        }
    }

    int steps = n + m + 2;

    map<pair<int, int>, int> mp;
    vector<pair<int, int>> vec;

    queue<tuple<int, int, int>> que;
    que.push({1, 1, a[1][1]});
    while (que.size())
    {
        auto [i, j, val] = que.front();
        que.pop();
        if (i + j == steps / 2)
        {
            mp[{i, val}]++;
        }
        else
        {
            if (i < n)
                que.push({i + 1, j, val ^ a[i + 1][j]});
            if (j < m)
                que.push({i, j + 1, val ^ a[i][j + 1]});
        }
    }

    que.push({n, m, a[n][m]});
    while (que.size())
    {
        auto [i, j, val] = que.front();
        que.pop();
        if (i + j == steps / 2)
        {
            vec.push_back({i, val ^ a[i][j]});
        }
        else
        {
            if (i > 1)
                que.push({i - 1, j, val ^ a[i - 1][j]});
            if (j > 1)
                que.push({i, j - 1, val ^ a[i][j - 1]});
        }
    }

    int ans = 0;

    for (auto [i, val] : vec)
    {
        ans += mp[{i, val ^ k}];
    }

    cout << ans << endl;
}