UVA10739 String to Palindrome【记忆化搜索+DP】

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
6
tanbirahmed
shahriarmanzoor
monirulhasan
syedmonowarhossain
sadrulhabibchowdhury
mohammadsajjadhossain
Sample Output
Case 1: 5
Case 2: 7
Case 3: 6
Case 4: 8
Case 5: 8
Case 6: 8

问题链接：UVA10739 String to Palindrome
问题简述：给定一个字符串，可以增加一个字母或者删除一个字母，或者替换任何一个字母，都算作一次操作，计算最少需要几次操作可以成为回文串。
问题分析：
三种操作是：删除一个字符，添加一个字符，替换一个字符。
本题提供两种解法，一是记忆化搜索，二是DP。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序（记忆化搜索）如下：

/* UVA10739 String to Palindrome */#include <bits/stdc++.h>using namespace std;const int N = 1000;
char s[N + 1];
int dp[N][N];int dfs(int left, int right)
{if(dp[left][right] != -1) return dp[left][right];if(left >= right) return dp[left][right] = 0;if(s[left] == s[right]) return dp[left][right] = dfs(left + 1, right - 1);else {  // s[left] != s[right]，删除字符dp[left][right] = dfs(left + 1, right); // 删除左边字符dp[left][right] = min(dp[left][right], dfs(left, right - 1)); // 删除右边字符dp[left][right] = min(dp[left][right], dfs(left + 1, right - 1));    // 删除左右字符return dp[left][right] += 1;}
}int main()
{int t;scanf("%d", &t);for(int k = 1; k <= t; k++) {scanf("%s", s);memset(dp, -1, sizeof(dp));printf("Case %d: %d\n", k, dfs(0, strlen(s) - 1));}return 0;
}

AC的C++语言程序（DP）如下：

/* UVA10739 String to Palindrome */#include <bits/stdc++.h>using namespace std;const int N = 1000;
char s[N + 1];
int dp[N][N];int main()
{int t;scanf("%d", &t);for(int k = 1; k <= t; k++) {scanf("%s", s);int len = strlen(s);for(int i = 0; s[i]; i++) dp[i][i] = 0;for(int i = len - 1; i >= 0; i--)for(int 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]);dp[i][j] = min(dp[i][j], dp[i + 1][j - 1]) + 1;}printf("Case %d: %d\n", k, dp[0][len - 1]);}return 0;
}

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UVA10739 String to Palindrome【记忆化搜索+DP】

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