hdu2476 String painter（区间dp）

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

参考: http://blog.csdn.net/libin56842/article/details/9708807

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8

typedef __int64 ll;

using namespace std;

#define N 105

int dp[N][N],ans[N];
char a[N],b[N];

int main()
{
   int i,j;
   while(~scanf("%s",a))
   {
   	  scanf("%s",b);
   	  int len=strlen(b);

   	  memset(dp,0,sizeof(dp));

   	  for(i=0;i<len;i++)
		dp[i][i]=1;

	  for(i=len-1;i>=0;i--)
	    for(j=i+1;j<len;j++)
		{
			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(i=0;i<len;i++)
	  {
	  	  ans[i]=dp[0][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\n",ans[len-1]);
   }
   return 0;
}