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jenningsloy318
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AlgorithmAndLeetCode:master
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youngyangyang04:master
Jul 26, 2024
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59 changes: 59 additions & 0 deletions
problems/kamacoder/0113.国际象棋.md
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| # 113.国际象棋 | ||
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| 广搜,但本题如果广搜枚举马和象的话会超时。 | ||
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| 广搜要只枚举马的走位,同时判断是否在对角巷直接走象 | ||
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| ```CPP | ||
| #include <iostream> | ||
| using namespace std; | ||
| const int N = 100005, mod = 1000000007; | ||
| using ll = long long; | ||
| int n, ans; | ||
| int dir[][2] = {{1, 2}, {1, -2}, {-1, 2}, {-1, -2}, {2, 1}, {2, -1}, {-2, -1}, {-2, 1}}; | ||
| int main() { | ||
| int x1, y1, x2, y2; | ||
| cin >> n; | ||
| while (n--) { | ||
| scanf("%d%d%d%d", &x1, &y1, &x2, &y2); | ||
| if (x1 == x2 && y1 == y2) { | ||
| cout << 0 << endl; | ||
| continue; | ||
| } | ||
| // 判断象走一步到达 | ||
| int d = abs(x1 - x2) - abs(y1 - y2); | ||
| if (!d) {cout << 1 << endl; continue;} | ||
| // 判断马走一步到达 | ||
| bool one = 0; | ||
| for (int i = 0; i < 8; ++i) { | ||
| int dx = x1 + dir[i][0], dy = y1 + dir[i][1]; | ||
| if (dx == x2 && dy == y2) { | ||
| cout << 1 << endl; | ||
| one = true; | ||
| break; | ||
| } | ||
| } | ||
| if (one) continue; | ||
| // 接下来为两步的逻辑, 象走两步或者马走一步,象走一步 | ||
| // 象直接两步可以到达,这个计算是不是同颜色的格子,象可以在两步到达所有同颜色的格子 | ||
| int d2 = abs(x1 - x2) + abs(y1 - y2); | ||
| if (d2 % 2 == 0) { | ||
| cout << 2 << endl; | ||
| continue; | ||
| } | ||
| // 接下来判断马 + 象的组合 | ||
| bool two = 0; | ||
| for (int i = 0; i < 8; ++i) { | ||
| int dx = x1 + dir[i][0], dy = y1 + dir[i][1]; | ||
| int d = abs(dx - x2) - abs(dy - y2); | ||
| if (!d) {cout << 2 << endl; two = true; break;} | ||
| } | ||
| if (two) continue; | ||
| // 剩下的格子全都是三步到达的 | ||
| cout << 3 << endl; | ||
| } | ||
| return 0; | ||
| } | ||
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| ``` |
132 changes: 132 additions & 0 deletions
problems/kamacoder/0121.小红的区间翻转.md
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| # 121. 小红的区间翻转 | ||
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| 比较暴力的方式,就是直接模拟, 枚举所有 区间,然后检查其翻转的情况。 | ||
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| 在检查翻转的时候,需要一些代码优化,否则容易超时。 | ||
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| ```CPP | ||
| #include <iostream> | ||
| #include <vector> | ||
| using namespace std; | ||
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| bool canTransform(const vector<int>& a, const vector<int>& b, int left, int right) { | ||
| // 提前检查翻转区间的值是否可以匹配 | ||
| for (int i = left, j = right; i <= right; i++, j--) { | ||
| if (a[i] != b[j]) { | ||
| return false; | ||
| } | ||
| } | ||
| // 检查翻转区间外的值是否匹配 | ||
| for (int i = 0; i < left; i++) { | ||
| if (a[i] != b[i]) { | ||
| return false; | ||
| } | ||
| } | ||
| for (int i = right + 1; i < a.size(); i++) { | ||
| if (a[i] != b[i]) { | ||
| return false; | ||
| } | ||
| } | ||
| return true; | ||
| } | ||
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| int main() { | ||
| int n; | ||
| cin >> n; | ||
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| vector<int> a(n); | ||
| vector<int> b(n); | ||
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| for (int i = 0; i < n; i++) { | ||
| cin >> a[i]; | ||
| } | ||
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| for (int i = 0; i < n; i++) { | ||
| cin >> b[i]; | ||
| } | ||
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| int count = 0; | ||
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| // 遍历所有可能的区间 | ||
| for (int left = 0; left < n; left++) { | ||
| for (int right = left; right < n; right++) { | ||
| // 检查翻转区间 [left, right] 后,a 是否可以变成 b | ||
| if (canTransform(a, b, left, right)) { | ||
| count++; | ||
| } | ||
| } | ||
| } | ||
| cout << count << endl; | ||
| return 0; | ||
| } | ||
| ``` | ||
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| 也可以事先计算好,最长公共前缀,和最长公共后缀。 | ||
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| 在公共前缀和公共后缀之间的部分进行翻转操作,这样我们可以减少很多不必要的翻转尝试。 | ||
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| 通过在公共前缀和后缀之间的部分,找到可以通过翻转使得 a 和 b 相等的区间。 | ||
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| 以下 为评论区 卡码网用户:码鬼的C++代码 | ||
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| ```CPP | ||
| #include <iostream> | ||
| #include <vector> | ||
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| using namespace std; | ||
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| int main() { | ||
| int n; | ||
| cin >> n; | ||
| vector<int> a(n), b(n); | ||
| for (int i = 0; i < n; i++) { | ||
| cin >> a[i]; | ||
| } | ||
| for (int i = 0; i < n; i++) { | ||
| cin >> b[i]; | ||
| } | ||
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| vector<int> prefix(n, 0), suffix(n, 0); | ||
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| // 计算前缀相等的位置 | ||
| int p = 0; | ||
| while (p < n && a[p] == b[p]) { | ||
| prefix[p] = 1; | ||
| p++; | ||
| } | ||
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| // 计算后缀相等的位置 | ||
| int s = n - 1; | ||
| while (s >= 0 && a[s] == b[s]) { | ||
| suffix[s] = 1; | ||
| s--; | ||
| } | ||
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| int count = 0; | ||
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| // 遍历所有可能的区间 | ||
| for (int i = 0; i < n - 1; i++) { | ||
| for (int j = i + 1; j < n; j++) { | ||
| // 判断前缀和后缀是否相等 | ||
| if ((i == 0 || prefix[i - 1] == 1) && (j == n - 1 || suffix[j + 1] == 1)) { | ||
| // 判断翻转后的子数组是否和目标数组相同 | ||
| bool is_palindrome = true; | ||
| for (int k = 0; k <= (j - i) / 2; k++) { | ||
| if (a[i + k] != b[j - k]) { | ||
| is_palindrome = false; | ||
| break; | ||
| } | ||
| } | ||
| if (is_palindrome) { | ||
| count++; | ||
| } | ||
| } | ||
| } | ||
| } | ||
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| cout << count << endl; | ||
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| return 0; | ||
| } | ||
| ``` |
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