String painter
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5659 Accepted Submission(s): 2695
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
#include<bits/stdc++.h> using namespace std; const int maxn=105; char t1[maxn],t2[maxn]; int dp[maxn][maxn],ans[maxn];//dp[i][j]表示区间i~j从空白串变为t2的最少操作次数 int main() { while(~scanf("%s%s",t1+1,t2+1)) { int L=strlen(t1+1); for(int i=L;i>=1;i--) { dp[i][i]=1; for(int j=i+1;j<=L;j++) { dp[i][j]=dp[i+1][j]+1; for(int k=i+1;k<=j;k++)//枚举i个字符和k一起变 { if(t2[i]==t2[k]) dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]); } } } for(int i=1;i<=L;i++) { if(t1[i]==t2[i]) ans[i]=ans[i-1]; else ans[i]=ans[i-1]+1;//假设第i个字符单独一次 for(int j=0;j<=i;j++)//枚举位置 ans[i]=min(ans[i],ans[j]+dp[j+1][i]); } printf("%d\n",ans[L]); } return 0; }