Problem H
String to Palindrome
Input: Standard Input
Output: Standard Output
Time Limit: 1 Second
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 Output for Sample Input
6 tanbirahmed shahriarmanzoor monirulhasan syedmonowarhossain sadrulhabibchowdhury mohammadsajjadhossain |
Case 1: 5 Case 2: 7 Case 3: 6 Case 4: 8 Case 5: 8 Case 6: 8 |
题意：给定一个字符串，可以进行添加，删除，和替换操作，求最少操作数使得该串变成回文串。
思路：i,j作为字符串的头尾。添加和删除操作其实是一样的。所以只需要考虑2种状态的转移了。
状态转移方程如果str[i] == str[j]则满足回文不用转换。dp[i][j] = dp[i + 1][j - 1]
如果不相等：dp[i][j] = min(min(dp[i + 1][j], dp[i][j - 1]), dp[i + 1][j - 1]) + 1
由于是递推。所以要从后往前。
代码：
#include <stdio.h>
#include <string.h>
int t, i, j, dp[1005][1005], len;
char sb[1005];
int min(int a, int b) {
return a < b ? a : b;
}
int main() {
scanf("%d%*c", &t);
int tt = 1;
while (t --) {
memset(dp, 0, sizeof(dp));
gets(sb);
len = strlen(sb);
for (i = len - 1; i >= 0; i --) {
for (j = i + 1; j < len; j ++) {
if (sb[i] == sb[j])
dp[i][j] = dp[i + 1][j - 1];
else
dp[i][j] = min(min(dp[i + 1][j], dp[i][j - 1]), dp[i + 1][j - 1]) + 1;
}
}
printf("Case %d: %d
", tt ++, dp[0][len - 1]);
}
return 0;
}