String painter
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5770 Accepted Submission(s): 2753
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
不会。。不知道在敲什么。。 AC了。。
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 105;
char s1[N],s2[N];
int dp[N][N];
int ans[N];
int main()
{
while(~scanf("%s%s",s1+1,s2+1))
{
int n = strlen(s1+1);
memset(dp,0,sizeof(dp));
for(int i = 1;i <= n;i ++)
{
for(int j = 1;j <= n;j ++)
{
dp[i][j] = j - i + 1;
}
}
for(int j = 1;j <= n;j ++)
{
for(int i = j;i >= 1;i --)
{
dp[i][j] = dp[i+1][j] + 1;
for(int k = i + 1;k <= j;k ++)
{
if(s2[i] == s2[k])
dp[i][j] = min(dp[i][j],dp[i+1][k]+dp[k+1][j]);
}
}
}
for(int i = 1;i <= n;i ++)
{
ans[i] = dp[1][i];
if(s1[i] == s2[i])
ans[i] = min(ans[i],ans[i-1]);
else
for(int j = 1;j < i;j ++)
{
ans[i] = min(ans[i],ans[j] + dp[j+1][i]);
}
}
cout << ans[n] << endl;
}
return 0;
}