hdu 2476 String painter

String painter
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1257 Accepted Submission(s): 532


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
zzzzzfzzzzz
abcdefedcba
abababababab
cdcdcdcdcdcd


Sample Output
6
7


Source
2008 Asia Regional Chengdu


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lcy

//46MS    292K    1034 B    C++
//区间DP,给两个字符串s1、s2,求将s1变成s2的最少操作步数的刷法
 
/*
第一步：假设A,B完全不等，没有位置上的字符相等，那么可以进行区间dp

dp[i][j]表示从i到j最少需要多少步

dp[i][j]=dp[i+1][j]+1;

dp[i][j]=MIN(dp[i][j],dp[i+j][k]+dp[k+1][j])  ---(i+1~j)

第二步：考虑s1和s2某些位置会相等

ans[i]=dp[0][i]  (if(s1[0]==s2[0]) ans[0]=0;)

ans[i]=ans[i-1]   s1[i]==s2[i]

ans[i]=Min(ans[i],ans[j]+dp[j+1][i]) 0<=j


ans[i]记录 0~i 的最优刷法步数 
*/

#include<stdio.h>
#include<string.h>
int dp[105][105];//dp[i][j]为i~j的刷法 
char s1[105],s2[105];
int ans[105],i,j,k,len;
int Min(int a,int b)
{
    return a<b?a:b;
}
int main(void)
{
    while(scanf("%s%s",s1,s2)!=EOF)
    {
        len=strlen(s1);
        memset(dp,0,sizeof(dp));
        for(int j=0;j<len;j++){
            for(int i=j;i>=0;i--){
                dp[i][j]=dp[i+1][j]+1; //每个单独刷 
                for(int k=i+1;k<=j;k++){  //i~j间最优刷法 
                    if(s2[i]==s2[k])
                        dp[i][j]=Min(dp[i][j],dp[i+1][k]+dp[k+1][j]); 
                }
                //printf("dp[%d][%d]=%d\n",i,j,dp[i][j]); 
            }
        }
        for(int i=0;i<len;i++) ans[i]=dp[0][i];
        if(s1[0]==s2[0]) ans[0]=0;
        for(int i=1;i<len;i++){ //类似dij的思想 
            if(s1[i]==s2[i]) ans[i]=ans[i-1];
            else{
                for(int j=0;j<i;j++)//区间最优 
                    ans[i]=Min(ans[i],ans[j]+dp[j+1][i]);
            }
        }
        printf("%d\n",ans[len-1]);
    }
    return 0;
}