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): 1117    Accepted Submission(s): 443

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

 

 

Recommend

lcy

 

 

 

 

一开始两个字符串很难搞。

其实可以先算由空白串直接变成str2.

用区间DP，可以求出dp[i][j].

然后再计算从str1变成str2.

//============================================================================

// Name        : HDU.cpp

// Author      : 

// Version     :

// Copyright   : Your copyright notice

// Description : Hello World in C++, Ansi-style

//============================================================================



#include <iostream>

#include <string.h>

#include <algorithm>

#include <stdio.h>

using namespace std;

const int MAXN=110;

int dp[MAXN][MAXN];

char str1[MAXN],str2[MAXN];

int ans[MAXN];

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    while(scanf("%s%s",str1,str2)==2)

    {

        int n=strlen(str1);

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

        for(int i=0;i<n;i++)

            for(int j=i;j<n;j++)

                dp[i][j]=j-i+1;

        //先直接DP求出从空白串变成str2

        for(int i=n-2;i>=0;i--)

            for(int j=i+1;j<n;j++)

            {

                dp[i][j]=dp[i+1][j]+1;

                for(int k=i+1;k<=j;k++)

                    if(str2[i]==str2[k])

                        dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k][j]);

            }

        for(int i=0;i<n;i++)

        {

            ans[i]=dp[0][i];

            if(str1[i]==str2[i])

            {

                if(i==0)ans[i]=0;

                else ans[i]=ans[i-1];

            }

            for(int j=0;j<i;j++)

                ans[i]=min(ans[i],ans[j]+dp[j+1][i]);

        }

        printf("%d\n",ans[n-1]);

    }

    return 0;

}