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1033 - Generating Palindromes

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
6 abcd aaaa abc aab abababaabababa pqrsabcdpqrs	Case 1: 3 Case 2: 0 Case 3: 2 Case 4: 1 Case 5: 0 Case 6: 9

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PROBLEM SETTER: MD. KAMRUZZAMAN

SPECIAL THANKS: JANE ALAM JAN (MODIFIED DESCRIPTION, DATASET)

题意：

给你一个字符串，问最少添加多少个字符，使原字符串变成回文串
思路：
回文串的性质是倒过来两者一样， 那么有这个性质可知， 设原始字符串为a，原始字符串颠倒过来的字符串为b，  那么求a和b的最长公共子序列， 字符串a的长度减去最长公共自序列的长度不就是最少添加的字符！

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
//typedef long long LL;
//typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int MAXN = 1e2 + 10;
int dp[MAXN][MAXN];
char str1[MAXN], str2[MAXN];
int main(int argc, char* argv[]) {
    int T;
    scanf("%d", &T);
    int Kcase = 0;
    while (T--) {
        scanf("%s", str1);
        int len = strlen(str1);
        memset(dp, 0, sizeof(dp));
        memset(str2, 0, sizeof(str2));
        for (int i = len - 1; i >= 0; i--) {
            str2[len - i - 1] = str1[i];
        }
        for (int i = 1; i <= len; i++) {
            for (int j = 1; j <= len; j++) {
                if (str1[i-1] == str2[j-1]) {
                    dp[i][j] = dp[i-1][j-1]+1;
                }
                else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
            }
        }
        printf("Case %d: %d
", ++Kcase, len - dp[len][len]);
    }
    return 0;
}