LightOJ 1033 Generating Palindromes（dp）
题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730#problem/A
题目：
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
6
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
题目大意：
给出一个字符串，求至少添加多少个元素使其变成回文串。
分析：
动态规划，（dp）,从起点和终点开始往中间进行dp
状态转移方程：
当s[i]==s[j]时，dp[i][j]=dp[i+1][j-1];
当s[i]！=s[j]时，dp[i][j]=min(dp[i+1][j],dp[i][j-1])+1;
代码：
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 7 char s[105]; 8 int dp[105][105]; 9 10 int min(int a,int b)//输出最小值 11 { 12 return a>b?b:a; 13 } 14 15 int main() 16 { 17 int t; 18 int n,m=1; 19 scanf("%d",&t); 20 while(t--) 21 { 22 scanf("%s",s+1);//从数组下标为1的位置开始 23 int n=strlen(s+1);//输入字符串的长度 24 memset(dp,0,sizeof(dp)); 25 for(int i=n;i>=1;i--)//从两边开始搜索 26 for(int j=i+1;j<=n;j++) 27 { 28 if(s[i]==s[j]) //s[i]与s[j]相等，搜索下一个 29 dp[i][j]=dp[i+1][j-1]; 30 else //s[i]与s[j]不相等，取一个最小值再加1 31 dp[i][j]=min(dp[i+1][j],dp[i][j-1])+1; 32 } 33 printf("Case %d: %d ",m++,dp[1][n]); 34 } 35 return 0; 36 }