hdu 2476 String painter

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1529    Accepted Submission(s): 680

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

2008 Asia Regional Chengdu

 

题意：

有两个字符串长度相同，现在有一个painter，一次可以把第一个字符串 中的一段区间内的所有字母都换成同一个字母（这个字母可以是任意一个），问最少执行多少次操作，才能将第一个字符串s1转换成第二个字符串s2.

思路：

先将空白串变为s2，dp[i][j]-将i~j刷为目标串所需要的步数。

dp[i][j]=dp[i][j-1]+1;  j单独刷

当s[j]==s[k]时，j可以借助刷k的时候一起刷，dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j-1]);

处理完后，我们就要看s1需要喷刷多少次到s2了，s1比空白串的优点在哪里呢？在于s1有与s2相同的字符，这些字符可以不用个刷本身就与s2相同，用res[i]记录1~i区间第二个字符串得出的喷刷次数，如果第一个字符串的i位置与第二个字符串的i位置相同，那么这个位置就不用喷刷了，res[i]=res[i-1],如果不相同，就要就要借助一个位于1~(i-1)区间内的变量来分割开，res[i]=min(res[i],res[j]+dp[j+1][i])。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 105
#define MAXN 100005
#define mod 100000000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int n,m,ans,cnt,tot;
char s1[maxn],s2[maxn];
int dp[maxn][maxn],res[maxn];

int main()
{
    int i,j,t,k,len;
    while(~scanf("%s%s",s1+1,s2+1))
    {
        memset(dp,0,sizeof(dp));
        n=strlen(s1+1);
        for(i=1;i<=n;i++)
        {
            dp[i][i]=1;
        }
        for(len=2;len<=n;len++)
        {
            for(i=1;i<=n;i++)
            {
                j=i+len-1;
                if(j>n) break ;
                dp[i][j]=dp[i][j-1]+1;
                for(k=i;k<j;k++)
                {
                    if(s2[k]==s2[j]) dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j-1]);
                }
            }
        }
        for(i=1;i<=n;i++)
        {
            res[i]=dp[1][i];
        }
        for(i=1;i<=n;i++)
        {
            if(s1[i]==s2[i]) res[i]=res[i-1];
            else
            {
                for(j=1;j<i;j++)
                {
                    res[i]=min(res[i],res[j]+dp[j+1][i]);
                }
            }
        }
        printf("%d\n",res[n]);
    }
    return 0;
}