poj1159Palindrome(LCS最长公共子序列变形)

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Palindrome

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 60920 Accepted: 21245

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 program which, given a string, determines the minimal number of characters to be inserted into the string in order to obtain a palindrome.

As an example, by inserting 2 characters, the string "Ab3bd" can be transformed into a palindrome ("dAb3bAd" or "Adb3bdA"). However, inserting fewer than 2 characters does not produce a palindrome.

Input

Your program is to read from standard input. The first line contains one integer: the length of the input string N, 3 <= N <= 5000. The second line contains one string with length N. The string is formed from uppercase letters from 'A' to 'Z', lowercase letters from 'a' to 'z' and digits from '0' to '9'. Uppercase and lowercase letters are to be considered distinct.

Output

Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.

Sample Input

5Ab3bd

Sample Output

题意：

给定一个长度为N的字符串。问你最少要插入多少个字符才能使它变成回文串。

分析：

一开始没看出来是LCS。假设原串和其倒过来的字符串的最长公共子序列长度为m，原串长度为n，则意味着原串中有m个字符是正着反着读顺序都一样的，那么只需要在合适的位置插入n-m个相应的字符，就可以使得整个串成为回文串。

先求得原串的逆串，通过DP求两者最长公共子序列的长度。自上而下，从左到右填充dp矩阵。

转译方程为：

if(a[i]==b[j]) dp[k][j]=dp[!k][j-1]+1;//当前位相同，截止当前位的LCS长度等于两个串都不考虑当前位时的LCS长度加一
else dp[k][j]=max(dp[!k][j],dp[k][j-1]);//否则，将其前面和上面方格中的较大值传递过来

由于题目给定空间有限，本题引入滚动数组（也可以用short代替int），并通过！运算符实现换行。和填充正方形矩阵的过程一样，填这个之前其前面的和上面的都已经填好了。

借一幅图加深理解：

（图片地址：http://blog.csdn.net/jianfpeng241241/article/details/51926420）

代码：

#include<iostream>#include<stdio.h>#include<math.h>#include<string.h>using namespace std;char a[5009],b[5009];int dp[2][5009];int main(){int n;while(~scanf("%d",&n)){scanf("%s",a);for(int i=0;i<n;i++){b[n-i-1]=a[i];}memset(dp,0,sizeof(dp));int k=0;for(int i=0;i<n;i++,k=!k){for(int j=0;j<n;j++){if(a[i]==b[j]) dp[k][j]=dp[!k][j-1]+1;else dp[k][j]=max(dp[!k][j],dp[k][j-1]);}}int ans=max(dp[0][n-1],dp[1][n-1]);printf("%d\n",n-ans);}return 0;}

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