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
题意:给定两个字符串a和b,求最少须要对a进行多少次操作,才干将a变成b。每次操作时将a中随意一段变成随意一个字母所组成的段。
#include<stdio.h>
#include<string.h>
int min(int a,int b)
{
return a>b?b:a;
}
int main()
{
int dp[105][105],ans[105];
char a[105],b[105];
while(scanf("%s%s",a,b)>0)
{
int len=strlen(a);
memset(dp,0,sizeof(dp));
for(int r=0;r<len;r++)
for(int i=0;i<len-r;i++)
{
int j=i+r;
dp[i][j]=dp[i+1][j]+1;
for(int k=i+1;k<=j;k++)
if(b[i]==b[k])
dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]);
}
for(int i=0;i<len; i++)
ans[i]=dp[0][i];
for(int i=0;i<len; i++)
if(a[i]==b[i])
{
if(i==0)ans[i]=0;
else ans[i]=ans[i-1];
}
else{
for(int k=0;k<i;k++)
ans[i]=min(ans[i],ans[k]+dp[k+1][i]);
}
printf("%d
",ans[len-1]);
}
}