题目 参考自博客:http://blog.csdn.net/u011498819/article/details/38356675 题意:查找这样的子回文字符串(未必连续,但是有从左向右的顺序)个数. 简单的区间dp,哎,以为很神奇的东西,其实也是dp,只是参数改为区间,没做过此类型的题,想不到用dp,以后就 知道了,若已经知道[0,i],推[0,i+1], 显然还要从i+1 处往回找,dp方程也简单: dp[j][i]=(dp[j+1][i]+dp[j][i-1]+10007-dp[j+1][…

题意 给定一个字符串,问有多少个回文子串(两个子串可以一样). 思路 注意到任意一个回文子序列收尾两个字符一定是相同的,于是可以区间dp,用dp[i][j]表示原字符串中[i,j]位置中出现的回文子序列的个数,有递推关系: dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]  (*) 如果i和j位置出现的字符相同,那么dp[i][j]可以由dp[i+1][j-1]中的子序列加上这两个字符构成回文子序列,也就是 dp[i][j]+=dp[i+1][j-1],注意…

Problem Description In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence <A, B, D> is a subsequence of <A,…

题目链接 题目大意 给你几个字符串 (1<len(s)<1000) ,要你求每个字符串的回文序列个数.对于10008取模. Solution 区间DP. 比较典型的例题. 状态定义: 令 \(f[i][j]\) 表示 \(i\) 到 \(j\) 的回文序列个数,\(s\) 为给出的字符串. 状态转移: \(s[i]\neq s[j]\) 那么此时 \(f[i][j]\) 即为\(f[i][j-1]\),\(f[i+1][j]\)之和. 但由于 \(i+1->j-1\)的我们明显重复统计了…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4632 题目大意:给你若干个字符串,回答每个字符串有多少个回文子序列(可以不连续的子串).解题思路: 设dp[i][j]为[i,j]的回文子序列数,那么得到状态转移方程: dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+MOD)%MOD if(str[i]==str[j]) dp[i][j]+=dp[i-1][j+1]+1 代码: #include<cstdi…

题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=4632 题意:求回文串子串的的个数. 思路:看转移方程就能理解了. dp[i][j] 表示区间i j 之间的回文子串的个数. 状态转移方程:dp[i][j]=dp[i][j-1]+dp[i+1][j]-dp[i+1][j-1];         if(str[i]==str[j]) dp[i][j]=dp[i][j]+dp[i+1][j-1]+1; 代码: #include<iostrea…

Palindrome subsequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65535 K (Java/Others) Total Submission(s): 2836 Accepted Submission(s): 1160 Problem Description In mathematics, a subsequence is a sequence that can be derived from a…

个人心得:动态规划真的是够烦人的,这题好不容易写出了转移方程,结果超时,然后看题解,为什么这些题目都是这样一步一步的 递推,在我看来就是懵逼的状态,还有那个背包也是,硬是从最大的V一直到0,而这个就是从把间距为1到ch.size()全部算出来,难道 这就是动态规划,无后效性,即每一步都是最优的状态,所以把所有状况全部解决然后就可以一步一步往后面推了??值得深思 网上题解: 分析:我们知道求添加最少的字母让其回文是经典dp问题,转化成LCS求解.这个是一个很明显的区间dp 我们定义dp [ i ]…

Problem Description Teacher Mai has n numbers a1,a2,⋯,anand n−1 operators("+", "-" or "*")op1,op2,⋯,opn−1, which are arranged in the form a1 op1 a2 op2 a3 ⋯ an.He wants to erase numbers one by one. In i-th round, there are n+…

Problem Description The TV shows such as You Are the One has been very popular. In order to meet the need of boys who are still single, TJUT hold the show itself. The show is hold in the Small hall, so it attract a lot of boys and girls. Now there ar…