NEFU OJ36 Magic Tree_alice and bob are playing with a magic tree this m-优快云博客

本文链接：https://blog.youkuaiyun.com/ustcyue/article/details/52667747

本文介绍了一个使用三维动态规划（DP）方法解决特定序列问题的C语言程序实例。该问题涉及两个DP数组，并通过递归的方式求解最长序列。代码中详细展示了状态转移方程的实现过程。

可以用dp,但有两个dp数组，而且是三维的，原谅我dp知识的匮乏。。

以后补上更具体说明

#include<stdio.h>
#include<string.h>

int maxs(int x, int y)
{
	if (x > y)
		return x;
	else
		return y;
}
int main()
{
	int dp[1002][32][2];
	int tree[1002];
	int mint, step, max;
	while (scanf("%d%d",&mint,&step) != EOF)
	{
		memset(dp, 0, sizeof(dp));
		max = -1;
		for (int i = 1; i <= mint;++i)
		{
			scanf("%d", &tree[i]);
		}
		for (int i = 1; i <= mint; ++i)
		{
			for (int j = 0; j <=step; --j)
			{
				if (tree[i] == 1)
				{
					dp[i][j][0] = dp[i-1][j][0] + 1;
					if (j <step)
					{
						dp[i][j][0] = maxs(dp[i][j][0], dp[i - 1][j+1][1] + 1);
					}
					dp[i][j][1] = dp[i-1][j][1];
					if (j <step)
					{
						dp[i][j][1] = maxs(dp[i][j][1], dp[i - 1][j+1][0]);
					}
					max = maxs(dp[i][j][0], max);
					max = maxs(dp[i][j][1], max);
				}
				else
				{
					dp[i][j][1] = dp[i - 1][j][1] + 1;
					if (j <step)
					{
						dp[i][j][1] = maxs(dp[i][j][1], dp[i - 1][j+1][0] + 1);
					}
					dp[i][j][0] = dp[i - 1][j][0];
					if (j <step)
					{
						dp[i][j][0] = maxs(dp[i][j][0], dp[i - 1][j+1][1]);
					}
					max = maxs(dp[i][j][0], max);
					max = maxs(dp[i][j][1], max);
				}
			}
		}
		printf("%d\n", max);
	}
	return 0;
}