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.
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
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 ",ans[n]); 35 } 36 return 0; 37 }