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In this problem you are asked to convert a string into a palindrome with minimum number of operations. The operations are described below:

Here you’d have the ultimate freedom. You are allowed to:

• Add any character at any position
• Remove any character from any position
• Replace any character at any position with another character

Every operation you do on the string would count for a unit cost. You’d have to keep that as low as possible.

For example, to convert “abccda” you would need at least two operations if we allowed you only to add characters. But when you have the option to replace any character you can do it with only one operation. We hope you’d be able to use this feature to your advantage.

Input

The input file contains several test cases. The first line of the input gives you the number of test cases, T (1≤T≤10). Then T test cases will follow, each in one line. The input for each test case consists of a string containing lower case letters only. You can safely assume that the length of this string will not exceed 1000 characters.

Output

For each set of input print the test case number first. Then print the minimum number of characters needed to turn the given string into a palindrome.

Sample Input                               Output for Sample Input

6

tanbirahmed

shahriarmanzoor

monirulhasan

syedmonowarhossain

sadrulhabibchowdhury

mohammadsajjadhossain

题意：有三种操作，每种操作花费一单位花费。问最少的花费

区间DP：求最小花费。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int INF=0x3f3f3f3f;
const int maxn=1100;
int dp[maxn][maxn];
char str[maxn];
int t;
int main()
{
     int cas=1;
     scanf("%d",&t);
     getchar();
     while(t--)
     {
         gets(str+1);
         int len=strlen(str+1);
         CLEAR(dp,INF);
         for(int i=1;i<=len;i++)
         {
             if(str[i]==str[i+1])
                dp[i][i+1]=0;
             dp[i][i]=0;
         }
         for(int l=1;l<=len;l++)
         {
             for(int i=1;i+l<=len;i++)
             {
                 int j=i+l;
                 if(str[i]==str[j])
                    dp[i][j]=min(dp[i][j],dp[i+1][j-1]);
                 else
                 {
                    dp[i][j]=min(dp[i+1][j],dp[i][j-1])+1;
                    dp[i][j]=min(dp[i][j],dp[i+1][j-1]+1);
                 }
             }
         }
         printf("Case %d: ",cas++);
         printf("%d\n",dp[1][len]);
     }
     return 0;
}

UVA 10739 String to Palindrome（DP)

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原文地址：http://blog.csdn.net/u013582254/article/details/41709911