abc 293 C

// DFS
#include <bits/stdc++.h>

using namespace std;

int h, w;
int a[11][11];

int dx[2] = {0, 1}, dy[2] = {1, 0};
set<int> s;

int dfs(int x, int y)
{
    int res = 0;
    s.insert(a[x][y]);
    if (x == h && y == w) return 1;


    for (int i = 0; i < 2; i ++) 
    {
        if (x + dx[i] <= h && y + dy[i] <= w && !s.count(a[x + dx[i]][y + dy[i]]))
        {
            res += dfs(x + dx[i], y + dy[i]);
            s.erase(a[x + dx[i]][y + dy[i]]);
        }
    }

    return res;
}

int main()
{
    cin >> h >> w;

    for (int i = 1; i <= h; i ++) 
        for (int j = 1; j <= w; j ++) 
            cin >> a[i][j];

    int ans = dfs(1, 1);
    s.erase(a[1][1]);

    cout << ans << endl;

    return 0;
}

abc 292 C

//排列组合的问题，a * b + c * d == n 的所有四元组
    int >> n;
    ll ans = 0;
    vector<ll> f(n + 1);
    for (int i = 1; i <= n; i ++) 
        for (int j = 1; j * i <= n; j ++)
            f[i * j] ++ ;

    for (int i = 1; i <= n; i ++) 
    {
        ans += f[i] * f[n - i];
    }

    cout << ans << endl;

abc 288 C

并查集中每次连接两个在同一个集合的点，会成一环，想要不成环，就要去掉这些边
答案就是这些边的个数