HDU_2476_String painter_(区间dp)

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3935    Accepted Submission(s): 1833

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

 

2017.3.18复习，感觉这道题的dp设计很巧妙，还是没有吃透。。。

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题意：给定两个等长字符串A、B，一次操作可以将A的任意一段连续的字符变为同一任意字符，问最少多少次操作可以将A变为B。

 

专门挑的区间dp的题，但是然并卵。。。没思路啊。。。看题解。。。

 

思路：dp[i][j]：将空串变为B[i-->j]所需要的最少次数。先求出所有dp[i][j]，即求出将空串变为B[i-->j]所需最少次数。那么根据dp数组求出最终答案就很容易:

1.ans[k]=dp[1][k]

 

2.if(A[k]==B[k])   ans[k]=ans[k-1];     (ans[0]==0)

　else　　ans[k]=min(ans[k],ans[t]+ans[t+1][k]);  (1=<t<k)

 

求dp[i][j]:

枚举起点i、终点j、遍历[i+1,j]区间k

for(int j=1;j<=len;j++) //end
{
　　for(int i=j;i>0;i--) //begin
　　{
　　　　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][k-1]+dp[k+1][j]);
　　　}
}

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

char A[105],B[105];
int dp[105][105];
int ans[105];

int main()
{
    while(scanf("%s%s",A+1,B+1)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        memset(ans,0,sizeof(ans));
        int len=strlen(B+1);
        for(int j=1;j<=len;j++)  //end
        {
            for(int i=j;i>0;i--)  //begin
            {
                dp[i][j]=dp[i+1][j]+1;
                for(int k=i+1;k<=j;k++)  //一定是从左往右找重的
                    if(B[i]==B[k])
                        dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]);
            }
        }
        for(int i=1;i<=len;i++)
        {
            ans[i]=dp[1][i];
            if(A[i]==B[i])
                ans[i]=ans[i-1];
            else
            {
                for(int j=1;j<i;j++)
                    ans[i]=min(ans[i],ans[j]+dp[j+1][i]);
            }
        }
        printf("%d\n",ans[len]);
    }
    return 0;
}