求多个串最长公共子序列,字典序最小输出.枚举剪枝+kmp.比较简单,我用find直接查找16ms #include<iostream> #include<string> #include<algorithm> using namespace std; string s[61]; int main() { int ta; cin>>ta; int n; while(ta--) { cin>>n; string ans; for(int i=0;i&…

POJ 3080 Blue Jeans (求最长公共字符串) Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated. As an IB…

Palindrome Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 56150   Accepted: 19398 Description A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a…

题意:有两个代表基因序列的字符串s1和s2,在两个基因序列中通过添加"-"来使得两个序列等长:其中每对基因匹配时会形成题中图片所示匹配值,求所能得到的总的最大匹配值. 题解:这题运用dp的解法是借用了求最长公共子序列的方法,,定义dp[i][j]代表s1以第i位结尾的串和s2以第j位结尾的串匹配时所能得到的最大匹配值:那么状态转移方程为:dp[i][j]=max( dp[i-1][j-1]+s1[i]和s2[j]的匹配值 , dp[i-1][j]+s1[i]和'-'的匹配值 , dp[…

解题思路:先注意到序列和串的区别,序列不需要连续,而串是需要连续的,先由样例abcfbc         abfcab画一个表格分析,用dp[i][j]储存当比较到s1[i],s2[j]时最长公共子序列的长度 a    b    f    c    a    b 0    0    0    0    0   0    0 a  0    1     1    1    1   1    1 b  0    1     2    2    2   2    2 c  0    1     2  …

题目链接Time Limit: 1000MS Memory Limit: 10000K Total Submissions: Accepted: Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = &…

Blue Jeans Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20966   Accepted: 9279 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousa…

Common Subsequence A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a s…

Palindrome [题目链接]Palindrome [题目类型]最长公共子序列 &题解: 你做的操作只能是插入字符,但是你要使最后palindrome,插入了之后就相当于抵消了,所以就和在这个串中删除最少的字符,使得它回文是一样的. 那么我们可以把这个串reverse,之后的串称为s2,找s2和s的最长公共子序列就好了,因为有了LCS,接着把其他的都删掉,就是一个回文串了,因为正着读和倒着读都一样 还有POJ居然能跑5000^2 我的923MS就跑完了,还是很快的嘛,当然这题还可以滚动数组,…

POJ 1458 最长公共子序列 题目大意:给出两个字符串,求出这样的一 个最长的公共子序列的长度:子序列 中的每个字符都能在两个原串中找到, 而且每个字符的先后顺序和原串中的 先后顺序一致. Sample Input : abcfbc abfcab programming contest abcd mnp Sample Output 4 2 0 分析: 输入两个串s1,s2, 设dp(i,j)表示: s1的左边i个字符形成的子串,与s2左边的j个 字符形成的子串的最长公共子序列的长度(i,j从…