Create 887.cpp

LeetCode-CN/887.cpp

@@ -0,0 +1,69 @@
+class Solution {
+public:
+	int superEggDrop(int K, int N) {
+		// dp[i][j] 用i个鸡蛋 在j步数内能走多少层
+		dp = new int*[K + 1];
+		for (int i = 1; i <= K; i++)
+		{
+			dp[i] = new int[N + 1];
+			dp[i][0] = 0;
+			dp[i][1] = 1;
+		}
+		for (int i = 1; i <= N; i++)
+		{
+			dp[1][i] = i;
+		}
+
+		for (int k = 2; k <= K; k++)
+		{
+			/* // O(n*n*k). TLE
+			for(int n=2;n<=N;n++)
+			{
+				int min_value = INT_MAX;
+				for (int j = 2; j <= n; j++)
+				{
+					min_value = min(min_value, max(dp[k - 1][j - 1], dp[k][n - j]) + 1);
+					printf("max(dp[%d][%d]=%d,dp[%d][%d]=%d)+1=%d\n", 
+						k - 1, j - 1, dp[k-1][j-1],
+						k, n - j, dp[k][n-j],
+						max(dp[k - 1][j - 1], dp[k][n - j]) + 1 );
+				}
+			}
+			*/
+			
+			for (int step = 2; step <= N; step++)
+			{
+				dp[k][step] = dp[k - 1][step - 1] // 在随便一层丢下去,碎了,则用k-1个鸡蛋,少一步能覆盖的数量表示这层下面有多少层
+					+ dp[k][step - 1] // 还是这层丢下去,没碎,那么就是k个鸡蛋,少一步能覆盖的数量表示这层上面有多少层
+					+ 1; // 就是这层自己
+			}
+		}
+
+		//print_table(K, N);
+
+		for (int step = 1; step <= N; step++)
+		{
+			if (dp[K][step] >= N) return step;
+		}
+	}
+	void print_table(int K, int N)
+	{
+		printf("DATA ");
+		for (int j = 1; j <= N; j++)
+		{
+			printf("%2d ", j);
+		}
+		printf("\n");
+		for (int i = 1; i <= K; i++)
+		{
+			printf("K=%2d ", i);
+			for (int j = 1; j <= N; j++)
+			{
+				printf("%2d ", dp[i][j]);
+			}
+			printf("\n");
+		}
+	}
+private:
+	int** dp;
+};