HDU 2476String painter 区间dp

M - String painter

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2476

Appoint description: System Crawler  (2016-04-25)

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 

 

思路是先求出从空白串转化到b串要花多少次

然后比较a串和b串相同元素的个数在优化答案

ACcode;

#include<iostream>#include<cstdio>#include<cstring>#define maxn 120using namespace std;int dp[maxn][maxn],ans[maxn];char a[maxn],b[maxn];int main(){    while(scanf("%s\n%s",a+1,b+1)!=EOF){        int n=strlen(a+1);        for(int i=0;i<=n;++i)dp[i][i]=1;        for(int l=2;l<=n;++l)        for(int i=1;i<=n-l+1;++i){            int j=i+l-1;            dp[i][j]=dp[i+1][j]+1;            for(int k=i+1;k<=j;++k)                if(b[k]==b[i])                    dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]);        }        ans[0]=0;        for(int i=1;i<=n;++i){            ans[i]=dp[1][i];            if(a[i]==b[i])ans[i]=ans[i-1];            for(int j=1;j<=i;++j)ans[i]=min(ans[i],dp[j+1][i]+ans[j]);        }        printf("%d\n",ans[n]);    }    return 0;}