hdu2476 String painter

Problem Description

There are two strings A and B with equal length. Both strings are made up of lower case letters. Now you have a powerful string painter. With the help of the painter, you can change a segment of characters of a string to any other character you want. That is, after using the painter, the segment is made up of only one kind of character. Now your task is to change A to B using string painter. What’s the minimum number of operations?

 

Input

Input contains multiple cases. Each case consists of two lines:
The first line contains string A.
The second line contains string B.
The length of both strings will not be greater than 100.

 

Output

A single line contains one integer representing the answer.

 

Sample Input

zzzzzfzzzzzabcdefedcbaababababababcdcdcdcdcdcd

 

Sample Output

67
题意：给出两个字符串a,b，将a串变为b串，每次可以将连续的一个子串改成任意的一个字母，问最少需要操作多少次。先考虑空串变为b串的情况：设dp[i][j]表示从i到j至少要变多少次，则有dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j])(i<=k<j)然后再考虑a串,因为a串中有和b串相同的字符，所以可能使操作数更少，设f[i]表示使a[1]~a[i]==b[1]~b[i]的最小步数，则有f[i]=min(f[j]+dp[j+1][i],dp[1][i],f[i-1](当a[i]==b[i]时) )，a串中的[j+1,i]可以看做一个空串。
#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;#define ll long long#define inf 999999999char s1[106],s2[106];int dp[106][106],f[106];int main(){    int n,m,i,j,len1,k,len,flag;    while(scanf("%s",s1+1)!=EOF)    {        scanf("%s",s2+1);        len1=strlen(s1+1);        for(i=1;i<len1;i++){            dp[i][i]=1;            if(s2[i]==s2[i+1]){                dp[i][i+1]=1;            }            else dp[i][i+1]=2;        }        dp[len1][len1]=1;        for(len=3;len<=len1;len++){            for(i=1;i+len-1<=len1;i++){                j=i+len-1;                dp[i][j]=len;                if(s2[i]==s2[j]){                    dp[i][j]=dp[i][j-1];continue;                }                for(k=i;k<j;k++){                    dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);                 }            }        }        if(s1[1]==s2[1]) f[1]=0;        else f[1]=1;        for(i=2;i<=len1;i++){            if(s1[i]==s2[i]){                f[i]=f[i-1];continue;            }            f[i]=dp[1][i];            for(k=1;k<i;k++){                if(s1[k]==s2[k]){                    f[i]=min(f[i],f[k]+dp[k+1][i]);                }            }        }        printf("%d\n",f[len1]);    }    return 0;}