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poj 1159 -- Palindrome
Palindrome
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 53250 | Accepted: 18396 |
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.
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
2
题目链接：Palindrome
思路：简单dp。头尾来一次最长公共子序列。用长度n-最长公共子序列即为答案。
1 /*====================================================================== 2 * Author : kevin 3 * Filename : Palindrome.cpp 4 * Creat time : 2014-09-25 10:31 5 * Description : 6 ========================================================================*/ 7 #include <iostream> 8 #include <algorithm> 9 #include <cstdio>10 #include <cstring>11 #include <queue>12 #include <cmath>13 #define clr(a,b) memset(a,b,sizeof(a))14 #define M 500515 using namespace std;16 char str1[M];17 short int dp[M][M];18 int main(int argc,char *argv[])19 {20 int n;21 while(scanf("%d",&n) != EOF){22 scanf("%s",str1);23 for(int i = 1; i <= n; i++){24 for(int j = 1; j <= n; j++){25 if(str1[i-1] == str1[n-j]){26 dp[i][j] = dp[i-1][j-1] + 1;27 }28 else{29 dp[i][j] = max(dp[i-1][j],dp[i][j-1]);30 }31 }32 }33 printf("%d\n",n - dp[n][n]);34 }35 return 0;36 }
poj 1159 -- Palindrome
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