light oj 1033 - Generating Palindromes (区间DP)

题目链接：http://lightoj.com/volume_showproblem.php?problem=1033

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

PROBLEM SETTER: MD. KAMRUZZAMAN SPECIAL THANKS: JANE ALAM JAN (MODIFIED DESCRIPTION, DATASET)

题目大意 ： 给你一串字符串，求最少添加多少个字符，使该串变成回文串，可以在任意地方添加

解析： 比赛时，完全懵逼了，不知道该从何入手，下去看大牛们的博客，才知道是区间DP， 把该

             字符串变成两个串，一个向头遍历，一个向尾部遍历，dp [ i ] [ j ] 表示长度为 j -  i + 1 的字符

             串需要添加的最小个数，如果是 s[ i ]  ==  s[ j ] ， dp[ i ]  [ j ] =  ;dp[i + 1] [ j - 1] ;

            不相等， dp [ i ] [ j ]  = min (dp[ i + 1] [ j ] , dp[ i ] [ j - 1] )  + 1;

具体见下面代码：

#include<iostream>
#include<algorithm>
#include<map>
#include<string>
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#define N  1009
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1000003;
int dp[N][N];
char s[N];

int main()
{
    int t, i, j, cnt = 0;
    cin >> t;
    while(t--)
    {
        scanf(" %s", s);
        int len = strlen(s);
        memset(dp, 0, sizeof(dp));
        for(i = len - 1; i >= 0; i--)
        {
            for(j = i + 1; j < len; j++)
            {
                if(s[i] == s[j]) dp[i][j] = dp[i + 1][j - 1];
                else dp[i][j] = min(dp[i + 1][j], dp[i][j - 1]) + 1;
            }
        }
        printf("Case %d: %d\n", ++cnt, dp[0][len - 1]);
    }
    return 0;
}