#10172. 「一本通 5.4 练习 1」涂抹果酱

https://loj.ac/s/1473671

LibreOJ/10172/10172.cpp

@@ -0,0 +1,107 @@
+#include <iostream>
+#include <string>
+#include <vector>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 10005,
+          M = 6;
+
+const int mod = 1e6;
+
+int n, m, k, t, a[M], f[N][245], ans;
+std::string ks;
+std::vector<std::string> r;
+
+void dfs(std::string s) {
+    if (s.size() >= m) {
+        if (s == ks) t = r.size();
+        r.push_back(s);
+
+        return;
+    }
+
+    if (s.back() == '1') dfs(s + '2'), dfs(s + '3');
+    else if (s.back() == '2') dfs(s + '1'), dfs(s + '3');
+    else dfs(s + '1'), dfs(s + '2');
+}
+
+bool check(std::string s) {
+    for (int i = 1; i < s.size(); i++) {
+        if (s[i] == s[i - 1]) return false;
+    }
+
+    return true;
+}
+
+bool check(std::string s1, std::string s2) {
+    for (int i = 0; i < s1.size(); i++) {
+        if (s1[i] == s2[i]) return false;
+    }
+
+    return true;
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    cin >> n >> m >> k;
+
+    for (int i = 1; i <= m; i++) {
+        cin >> a[i];
+        ks.push_back(a[i] + '0');
+    }
+
+    if (!check(ks)) {
+        cout << 0 << endl;
+
+        exit(0);
+    }
+
+    dfs("1");
+    dfs("2");
+    dfs("3");
+
+    if (k == 1) {
+        f[1][t] = 1;
+    } else {
+        for (int i = 0; i < r.size(); i++) {
+            f[1][i] = 1;
+        }
+    }
+
+    for (int i = 2; i <= n; i++) {
+        if (i == k) {
+            for (int j = 0; j < r.size(); j++) {
+                if (check(r[t], r[j])) {
+                    f[i][t] = (f[i][t] + f[i - 1][j]) % mod;
+                }
+            }
+        } else {
+            for (int j = 0; j < r.size(); j++) {
+                if (i - 1 == k) {
+                    if (check(r[t], r[j])) {
+                        f[i][j] = (f[i][j] + f[i - 1][t]) % mod;
+                    }
+                } else {
+                    for (int k = 0; k < r.size(); k++) {
+                        if (check(r[j], r[k])) {
+                            f[i][j] = (f[i][j] + f[i - 1][k]) % mod;
+                        }
+                    }
+                }
+            }
+        }
+    }
+
+    for (int i = 0; i < r.size(); i++) {
+        ans = (ans + f[n][i]) % mod;
+    }
+
+    cout << ans << endl;
+
+    return 0;
+}