String to Palindrome+uva+回文串dp

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

解决方案：题意是这样的，给一个字符串，求可以在任何位置加一个字符、减掉一个字符、或改变一个字符，让其成为回文串，求得是最小操作数。回文串dp,dp[i][j]表示字符串i到j变成回文所需的最小操作数，若text[i]==text[j],则dp[i][j]=dp[i+1][j-1]，若text[i]!=text[j]，则情况有几种，删掉text[i]或text[j]，改变text[i]或text[j]让其相等，或在i或j的位置插一个字符然后构成text[i]==text[j]，

所以dp[i][j]=min{dp[i+1][j-1]+1,dp[i+1][j]+1,dp[i][j-1]+1};边界条件为dp[i][i]=0,若text[i]==text[i+1],dp[i][i+1]=0;

code:

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1005;
int dp[maxn][maxn];
char text[maxn];
int main()
{
    int t,k=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",text+1);
        int len=strlen(text+1);
        memset(dp,0x3f,sizeof(dp));
        for(int i=1; i<=len; i++)
        {
            if(text[i]==text[i+1]){
                dp[i][i+1]=0;
            }
            dp[i][i]=0;
        }
        for(int i=2; i<=len; i++)
        {
            for(int j=1; j+i<=len; j++)
            {
                if(text[j]==text[j+i])
                {
                    dp[j][j+i]=dp[j+1][j+i-1];
                }
                else
                {

                    int temp=dp[j+1][j+i-1]+1;
                    dp[j][j+i]=min(dp[j+1][j+i]+1,dp[j][j+i-1]+1);
                    dp[j][j+i]=min(dp[j][j+i],temp);

                }
            }
        }
        printf("Case %d: %d\n",++k,dp[1][len]);
    }
    return 0;
}