解题报告PKU1159 最长公共字串 回文串 DP

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 31491		Accepted: 10509

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

5

Ab3bd

Sample Output

2

 

 

 

题目要求计算在给定的串中添加至少几个字符能使原来的串构成回文串 , 这题可以利用最长公共字串求解 , 设给定长度为 n 的字符串 s, 其反向串为 ts, 则 s 和 ts 的最长公共字串就是能够构成回文串的部分 , 用 n 减去最长公共字串的长度就是最少需要添加的字符数 .

 

代码如下 :

 

 

 

#include<stdio.h> #include<string.h> short res[5005][5005]; //用int会超内存 int main() { int i,j,n;char s[5005],ts[5005]; while (scanf("%d",&n)!=EOF) { gets(s); gets(s); memset(res,0,sizeof(res)); for (i=0;i<n;++i) ts[i]=s[n-1-i]; for (i=0;i<n;++i) for (j=0;j<n;++j) if (s[i]==ts[j]) res[i+1][j+1]=res[i][j]+1; else res[i+1][j+1]=res[i][j+1]>res[i+1][j]?res[i][j+1]:res[i+1][j]; printf("%d/n",n-res[n][n]); } return 0; }