LIS（最长递增子序列）和LCS(最长公共子序列）的总结

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LIS（最长递增子序列）和LCS(最长公共子序列）的总结

最长公共子序列（LCS）:O(n^2)

两个for循环让两个字符串按位的匹配:i in range(1, len1) j in range(1, len2)
s1[i - 1] == s2[j - 1], dp[i][j] = dp[i - 1][j -1] + 1;
s1[i - 1] != s2[j - 1], dp[i][j] = max (dp[i - 1][j], dp[i][j - 1]);
初始化：dp[i][0] = dp[0][j] = 0;

伪代码：

    dp[maxn1][maxn2];s1[maxn1],s2[maxn2];p[maxn1][maxn2][2];//initfor i in range(0, len1):dp[i][0] = 0;else:;for i in range(0, len2):dp[0][i] = 0;else:;for i in range(1, len1):for j in range(1, len2):if s1[i] == s2[j]:dp[i][j] = dp[i - 1][j - 1] + 1;p[i][j][0] = i - 1;p[i][j][1] = j - 1;else:if dp[i - 1][j] > dp[i][j - 1]:dp[i][j] = dp[i - 1][j];p[i][j][0] = i - 1;p[i][j][1] = j;else:dp[i][j] = dp[i][j - 1];p[i][j][0] = i;p[i][j][1] = j - 1;else:;else:;return dp[len1][len2];//path 非递归function print_path(len1, len2):if (dp[len1][len2] == 0)return;printf_path(p[len1][len2][0], p[len1][len2][1]);if s1[len1] == s2[len2]:printf:s1[len1];end function;

题目：UVA - 531Compromise UVA - 10066The Twin Towers UVA - 10192Vacation
uva10405 - Longest Common Subsequence

最长递增子序列（LIS）：O(n^2)

从左到右的求前i长度的序列的最长递增子序列的长度，状态转移方程：
dp[i] = Max（dp[j] + 1）;i in range(1, len); j in range(1, i - 1);

伪代码

    s[maxn],dp[maxn];for i in range(1, len):dp[i] = 1;int maxlen = 1;for i in range(2, len):for j range(1, i - 1):if s[i] > s[j]:dp[i] = Max(dp[i], dp[j] + 1);else:maxlen = max(maxlen, dp[i]);else:;return maxlen;//path递归function print_path(maxlen):if maxlen == 0:return;for i in range(1, len):if dp[i] == maxlen:print_path(maxlen - 1);printf:s[i];end function;

题目： UVA - 10599Robots(II)

最长递增子序列O(n * logn)

还是从左往右的求前i长度的序列的最长递增子序列长度，但是再确定dp[j]最大值的时候还要用一层循环来查找，这样比较低效.如果把前面的i长度序列出现的最长递增子序列储存起来，那么查找的时候用二分就可以做到O(logn)的复杂度。
用一个LIS数组来储蓄前i序列的最长递增子序列，查找第i个数字的时候，如果num[i] > LIS[top], 那么LIS[++top] = num[i]; dp[i] = top;如果num[i] == LIS[top],那么dp[i] = top; 如果num[i] < LIS[top], 那么二分查找到某个等于或者大于num[i]的最接近的值的位置（第k个），dp[i] = k - 1; LIS[k] = num[i];

题目：UVA - 10534Wavio Sequence

伪代码

    dp[maxn], LIS[maxn], s[maxn];top = 0;LIS[top++] = s[1];int maxlen = 1;for i in range(2, len):if s[i] > LIS[top]:LIS[++top] = s[i];dp[i] = top + 1;else if s[i] == LIS[top]:dp[i] = top + 1;else:k = lower_bound(LIS.begin(), LIS.end(), s[i]) - LIS.beign();LIS[k] = s[i];dp[i] = k + 1;maxlen = max(maxlen, dp[i]);else:;return maxlen;