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HDU 2476 String Painter

2024-07-29 19:52:49   220人阅读

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String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)


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

 

67

 

 
Source

 

2008 Asia Regional Chengdu

 

比较好的区间dp，可以先求空串到目标串的最少操作数

然后再对原串进行dp

 1 #include<set> 2 #include<queue> 3 #include<vector> 4 #include<cstdio> 5 #include<cstring> 6 #include<iostream> 7 #include<algorithm> 8 using namespace std; 9 const int MAXN=110;10 #define For(i,n) for(int i=1;i<=n;i++)11 #define Rep(i,l,r) for(int i=l;i<=r;i++)12 #define Down(i,r,l) for(int i=r;i>=l;i--)13 int dp[MAXN][MAXN];14 char str1[MAXN],str2[MAXN];15 int ans[MAXN];16 int main(){17     while(~scanf("%s%s",str1+1,str2+1)){18         int n=strlen(str1+1);19         memset(dp,0,sizeof(dp));20         For(i,n)21             Rep(j,i,n) dp[i][j]=j-i+1;22         Down(i,n-1,1)23             Rep(j,i+1,n){24                 dp[i][j]=dp[i+1][j]+1;25                 Rep(k,i+1,j)26                     if(str2[i]==str2[k])27                         dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k][j]);28             }29         For(i,n){30             ans[i]=dp[1][i];31             if(str1[i]==str2[i])  ans[i]=ans[i-1];32             For(j,i)              ans[i]=min(ans[i],ans[j]+dp[j+1][i]);33         }34         printf("%d\n",ans[n]);35     }36     return 0;37 }
Codes

 

HDU 2476 String Painter


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