String painter
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4108 Accepted Submission(s): 1915
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
明天来解释 我想洗澡睡觉了
1 #include<iostream>
2 #include<cstdio>
3 #include<cstdlib>
4 #include<cctype>
5 #include<cmath>
6 #include<cstring>
7 #include<map>
8 #include<stack>
9 #include<set>
10 #include<vector>
11 #include<algorithm>
12 #include<string.h>
13 typedef long long ll;
14 typedef unsigned long long LL;
15 using namespace std;
16 const int INF=0x3f3f3f3f;
17 const double eps=0.0000000001;
18 const int N=1000+10;
19 char a[N],b[N];
20 int dp[N][N];
21 int DP[N];
22 int main(){
23 while(gets(a)){
24 gets(b);
25 int len=strlen(a);
26 memset(dp,INF,sizeof(dp));
27 for(int i=0;i<len;i++)dp[i][i]=1;
28 for(int t=1;t<len;t++){
29 for(int i=0;i+t<len;i++){
30 int j=i+t;
31 if(b[i]==b[j])dp[i][j]=dp[i][j-1];
32 else{
33 for(int k=i;k<j;k++)
34 dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);
35 }
36 }
37 }
38 for(int i=0;i<len;i++){
39 if(i==0&&a[i]==b[i])dp[0][i]=0;
40 else if(a[i]==b[i])dp[0][i]=dp[0][i-1];
41 for(int j=0;j<i;j++)
42 dp[0][i]=min(dp[0][i],dp[0][j]+dp[j+1][i]);
43 }
44 cout<<dp[0][len-1]<<endl;
45 }
46 }