题目大意
从(1,1)格子开始不重复地经过N*N个格子(包括(1,1)和(N,1))到达(N,1),求路径方案数.
题解
搜索剪枝.
剪枝:
1.当把网格分成两部分时剪枝.
2.走到角的附近时一定走角(N,1)除外.
代码
/* TASK:betsy LANG:C++ */ #include <cstdio> #include <cstring> #include <iostream> using namespace std; const int d[2][4] = {{-1, 0, 1, 0}, {0, -1, 0, 1}}; int n, f[10][10]; bool vis[10][10]; bool check(int x, int y) { if (!vis[x-1][y] && !vis[x+1][y] && vis[x][y-1] && vis[x][y+1]) return true; if (!vis[x][y-1] && !vis[x][y+1] && vis[x-1][y] && vis[x+1][y]) return true; return false; } long long dfs(int x, int y, int dep) { if (x == n && y == 1) { if (dep == n * n) return 1; return 0; } if (check(x, y)) return 0; int msc = 0, k; for (int i = 0; i < 4; ++i) { int tx = x + d[0][i], ty = y + d[1][i]; if (vis[tx][ty]) { f[tx][ty]--; if (f[tx][ty] == 1) msc++, k = i; } } vis[x][y] = false; long long res = 0; if (msc <= 1) { if (msc == 0) for (int i = 0; i < 4; ++i) { int tx = x + d[0][i], ty = y + d[1][i]; if (vis[tx][ty]) res += dfs(tx, ty, dep + 1); } else res = dfs(x+d[0][k], y+d[1][k], dep+1); } for (int i = 0; i < 4; ++i) { int tx = x + d[0][i], ty = y + d[1][i]; if (vis[tx][ty]) f[tx][ty]++; } vis[x][y] = true; return res; } int main() { freopen("betsy.in", "r", stdin); freopen("betsy.out", "w", stdout); scanf("%d", &n); memset(vis, true, sizeof(vis)); for (int i = 0; i <= n; ++i) vis[0][i] = vis[i][0] = vis[n+1][i] = vis[i][n+1] = false; for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) f[i][j] = 4; for (int i = 1; i <= n; ++i) f[i][1] = f[1][i] = f[i][n] = f[n][i] = 3; f[n][1] = 10; f[1][1] = f[1][n] = f[n][n] = 2; cout << dfs(1, 1, 1) << endl; return 0; }