Light OJ 1033 .5THproblem A. DP 最长回文子序列

Description

By definition palindrome is a string which is not changed when reversed. "MADAM" is a nice example of palindrome. It is an easy job to test whether a given string is a palindrome or not. But it may not be so easy to generate a palindrome.

Here we will make a palindrome generator which will take an input string and return a palindrome. You can easily verify that for a string of length n, no more than (n - 1) characters are required to make it a palindrome. Consider "abcd" and its palindrome "abcdcba" or "abc" and its palindrome "abcba". But life is not so easy for programmers!! We always want optimal cost. And you have to find the minimum number of characters required to make a given string to a palindrome if you are only allowed to insert characters at any position of the string.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains a string of lowercase letters denoting the string for which we want to generate a palindrome. You may safely assume that the length of the string will be positive and no more than 100.

Output

For each case, print the case number and the minimum number of characters required to make string to a palindrome.

Sample Input

abcd

aaaa

abc

aab

abababaabababa

pqrsabcdpqrs

Sample Output

Case 1: 3

Case 2: 0

Case 3: 2

Case 4: 1

Case 5: 0

Case 6: 9

分析：

dp[i][j]表示从i到j最少要添加的字符个数

if(a[i]==a[j])
                dp[i][j]=dp[i+1][j-1]; //dp等于上一个状态的dp
            else
                dp[i][j]=min(dp[i+1][j],dp[i][j-1])+1; //取两种划分的最小值

代码及分析如下：

 1 //dp[i][j]表示从i到j最少要添加的字符个数
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<string>
 6 #include<algorithm>
 7 using namespace std;
 8 int dp[105][105];
 9 string a;
10 int min(int x,int y)
11 {
12     if(x<=y)
13         return x;
14     else
15         return y;
16 }
17 
18 int main()
19 {
20     int t,len,i,j,Case=1;
21     cin>>t;
22     while(t--)
23     {
24         cin>>a;
25         len=a.size();
26         memset(dp,0,sizeof(dp)); 
27         for(i=len-2;i>=0;i--)    //注意i从倒数第二个位置开始
28             for(j=i+1;j<=len-1;j++)  //j<=len-1而不是j<=len-2
29            {
30             if(a[i]==a[j])
31                 dp[i][j]=dp[i+1][j-1];  //dp等于上一个状态的dp
32             else
33                 dp[i][j]=min(dp[i+1][j],dp[i][j-1])+1;  //取两种划分的最小值
34 
35            }
36         printf("Case %d: %d
",Case++,dp[0][len-1]);
37     }
38 
39     return 0;
40 }

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