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288 lines
8.6 KiB
Python
Executable file
288 lines
8.6 KiB
Python
Executable file
"""#
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### 谜题描述
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You have a rectangular n × m-cell board. Some cells are already painted some of k colors. You need to paint each uncolored cell one of the k colors so that any path from the upper left square to the lower right one doesn't contain any two cells of the same color. The path can go only along side-adjacent cells and can only go down or right.
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Print the number of possible paintings modulo 1000000007 (109 + 7).
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Input
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The first line contains three integers n, m, k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 10). The next n lines contain m integers each — the board. The first of them contains m uppermost cells of the board from the left to the right and the second one contains m cells from the second uppermost row and so on. If a number in a line equals 0, then the corresponding cell isn't painted. Otherwise, this number represents the initial color of the board cell — an integer from 1 to k.
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Consider all colors numbered from 1 to k in some manner.
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Output
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Print the number of possible paintings modulo 1000000007 (109 + 7).
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Examples
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Input
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2 2 4
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0 0
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0 0
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Output
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48
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Input
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2 2 4
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1 2
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2 1
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Output
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0
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Input
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5 6 10
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0 0 0 0 0 0
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0 0 0 0 0 0
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0 0 0 0 0 0
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0 0 0 0 0 0
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0 0 0 0 0 0
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Output
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3628800
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Input
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2 6 10
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1 2 3 4 5 6
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0 0 0 0 0 0
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Output
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4096
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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#include <bits/stdc++.h>
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using namespace std;
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long long n, m, k, i, j, lim[9][9], lim2[9][9], a[9][9], s, mod = 1e9 + 7;
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vector<long long> v;
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long long dfs(long long x, long long y, long long cnt) {
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if (x > n) return 1;
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if (y > m) return dfs(x + 1, 1, cnt);
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long long ans = 0, i;
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lim[x][y] = lim[x - 1][y] | lim[x][y - 1];
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for (i = 0; i < k; i++) {
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if (a[x][y] != -1 && a[x][y] != i) continue;
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if (lim[x][y] & (1 << i)) continue;
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if (lim2[x][y] & (1 << i)) continue;
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if (!((1 << i) & s)) {
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if (a[x][y] == -1 && v[cnt] < i) continue;
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}
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lim[x][y] ^= (1 << i);
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long long s =
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dfs(x, y + 1, min((long long)((int)v.size() - 1), cnt + (v[cnt] == i)));
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if (i == v[cnt] && a[x][y] == -1)
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(ans += s * (v.size() - cnt)) %= mod;
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else
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(ans += s) %= mod;
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lim[x][y] ^= (1 << i);
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}
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return ans;
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}
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int main() {
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scanf(\"%I64d%I64d%I64d\", &n, &m, &k);
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if (n + m > 11 || n + m - 1 > k) {
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cout << 0;
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return 0;
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}
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for (i = 1; i <= n; i++) {
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for (j = 1; j <= m; j++) {
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scanf(\"%I64d\", &a[i][j]);
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a[i][j]--;
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if (a[i][j] >= 0) s |= (1 << a[i][j]);
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if (a[i][j] >= 0) lim2[i - 1][j] |= (1 << a[i][j]);
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if (a[i][j] >= 0) lim2[i][j - 1] |= (1 << a[i][j]);
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}
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}
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for (i = n; i >= 1; i--) {
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for (j = m; j >= 1; j--) {
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if (a[i][j] >= 0 && ((lim2[i][j] >> a[i][j]) & 1)) {
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cout << 0;
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return 0;
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}
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lim2[i][j] |= lim2[i + 1][j] | lim2[i][j + 1];
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}
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}
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for (i = 0; i < k; i++)
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if ((s >> i) % 2 == 0) v.push_back(i);
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if (v.size() == 0) {
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cout << 1;
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return 0;
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}
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cout << dfs(1, 1, 0);
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return 0;
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}
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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import re
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import random
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from bootcamp import Basebootcamp
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MOD = 10**9 + 7
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class Ddistinctpathsbootcamp(Basebootcamp):
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def __init__(self, max_n=5, max_m=5, max_k=10):
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self.max_n = max_n
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self.max_m = max_m
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self.max_k = max_k
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def generate_valid_full_grid(self, n, m, k):
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grid = [[0 for _ in range(m)] for _ in range(n)]
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for i in range(n):
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for j in range(m):
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used = set()
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if i > 0:
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used.add(grid[i-1][j])
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if j > 0:
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used.add(grid[i][j-1])
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available = [c for c in range(1, k+1) if c not in used]
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if not available:
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return None
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grid[i][j] = min(available)
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return grid
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def case_generator(self):
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n, m = 0, 0
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for _ in range(100): # Retry to get valid n, m
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n = random.randint(1, self.max_n)
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m = random.randint(1, self.max_m)
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if n + m <= 11:
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break
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k = random.randint(max(n + m -1, 1), self.max_k)
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grid_full = self.generate_valid_full_grid(n, m, k)
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if not grid_full:
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grid_initial = [[0]*m for _ in range(n)]
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else:
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grid_initial = []
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for row in grid_full:
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new_row = []
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for val in row:
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if random.random() < 0.3: # 30% chance to retain color
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new_row.append(val)
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else:
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new_row.append(0)
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grid_initial.append(new_row)
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# Compute expected solution
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expected = self.compute_solution(n, m, k, grid_initial)
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return {
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'n': n,
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'm': m,
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'k': k,
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'grid': grid_initial,
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'expected': expected
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}
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@staticmethod
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def compute_solution(n, m, k, grid):
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if n + m > 11 or (n + m - 1) > k:
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return 0
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grid = [[cell-1 if cell !=0 else -1 for cell in row] for row in grid]
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a = [[-1]*(m+2) for _ in range(n+2)]
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for i in range(n):
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for j in range(m):
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a[i+1][j+1] = grid[i][j] if grid[i][j] != -1 else -1
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lim2 = [[0]*(m+2) for _ in range(n+2)]
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s = 0
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for i in range(1, n+1):
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for j in range(1, m+1):
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if a[i][j] != -1:
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s |= 1 << a[i][j]
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if i < n:
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lim2[i][j] |= 1 << a[i][j]
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if j < m:
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lim2[i][j] |= 1 << a[i][j]
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for i in range(n, 0, -1):
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for j in range(m, 0, -1):
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lim2[i][j] |= lim2[i+1][j] | lim2[i][j+1]
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if a[i][j] != -1 and (lim2[i][j] & (1 << a[i][j])):
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return 0
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v = []
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for color in range(k):
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if not (s & (1 << color)):
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v.append(color)
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if not v:
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return 1
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# DFS to compute answer
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memo = {}
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def dfs(x, y, cnt, lim):
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if x > n:
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return 1
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if y > m:
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return dfs(x+1, 1, cnt, lim)
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key = (x, y, cnt, tuple(map(tuple, lim)))
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if key in memo:
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return memo[key]
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current_lim = lim[x-1][y] | lim[x][y-1]
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total = 0
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for color in range(k):
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if a[x][y] != -1 and color != a[x][y]:
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continue
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if current_lim & (1 << color):
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continue
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if lim2[x][y] & (1 << color):
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continue
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if not (s & (1 << color)):
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if a[x][y] == -1 and (cnt >= len(v) or color > v[cnt]):
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continue
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new_lim = [row[:] for row in lim]
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new_lim[x][y] = current_lim | (1 << color)
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new_cnt = cnt
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if a[x][y] == -1 and not (s & (1 << color)):
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if color == v[cnt]:
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new_cnt = min(len(v)-1, cnt + 1)
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res = dfs(x, y+1, new_cnt, new_lim)
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if a[x][y] == -1 and color in v and cnt < len(v) and color == v[cnt]:
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total = (total + res * (len(v) - cnt)) % MOD
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else:
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total = (total + res) % MOD
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memo[key] = total
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return total
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lim_init = [[0]*(m+2) for _ in range(n+2)]
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result = dfs(1, 1, 0, lim_init)
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return result
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@staticmethod
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def prompt_func(question_case):
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grid = question_case['grid']
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n = question_case['n']
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m = question_case['m']
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k = question_case['k']
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grid_str = '\n'.join(' '.join(map(str, row)) for row in grid)
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return f"""You are given a {n}x{m} grid. Some cells are colored (1 to {k}) and others are 0 (uncolored). Your task is to count the number of ways to color all 0-cells such that every path from the top-left to the bottom-right (moving only right or down) has all distinct colors. Output the count modulo 1e9+7.
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Input:
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{n} {m} {k}
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{grid_str}
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The output must be the number of valid ways modulo 1e9+7. Place your final answer within [answer] tags, e.g., [answer]12345[/answer]."""
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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return int(matches[-1].strip()) if matches else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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return solution == identity['expected']
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