刷题总结——String painter（hdu2476）

题目：

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

题解：

我们先预处理出一个f[i][j]，表示一个空白的串i-j部分如果要涂成第二个串一样且i率先涂上相同颜色（可以连着涂··只要保证i先匹配上）需要的最少次数···然后再更新ans[i]即可··具体看代码

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<cctype>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
const int N=105;
int n,f[N][N],ans[N];
char s[N],t[N];
int main()
{
 // freopen("a.in","r",stdin);
  while(~scanf("%s%s",s+1,t+1))
  { 
    n=strlen(s+1);
    memset(f,inf,sizeof(f));
    memset(ans,inf,sizeof(ans));
    ans[0]=0;
    for(int i=1;i<=n;i++) f[i][i]=1;
    for(int i=1;i<n;i++)
      if(t[i]==t[i+1])  f[i][i+1]=1;
      else f[i][i+1]=2;
    for(int i=n-2;i>=1;i--)
      for(int j=i+2;j<=n;j++) 
      {
        f[i][j]=f[i+1][j]+1;  
        for(int k=i+1;k<=j;k++)
          if(t[i]==t[k])  f[i][j]=min(f[i][j],f[i+1][k-1]+f[k][j]);//在涂上位置时顺便涂上i位置 
      }
    
    for(int i=1;i<=n;i++)
    {  
      if(s[i]==t[i]) ans[i]=ans[i-1];  //如果这个位置已经相等就不涂 
      else
        for(int j=0;j<i;j++)
          ans[i]=min(ans[i],ans[j]+f[j+1][i]);   //否则找到最优解 
    }
    cout<<ans[n]<<endl;
  }
  return 0;
}