HDU 2476 String painter

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1312    Accepted Submission(s): 554

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

 

在程序编写过程中有很多问题，需要理解一下。

      for(len=2;len<=n;len++)//枚举长度
        {
            for(j=len-1;j<n;j++)//枚举终点
            {
                i=j-len+1;//起点
                dp[i][j]=dp[i+1][j]+1;
                for(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]);
                    }
                }
            }
        } 

比较

for(i=0;i<n;i++)
        {
            for(j=i+1;j<n;j++)
            {
                for(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]);
                    }
                }
            }
        }

前者第一个枚举长度，紧接着是枚举终点，和起始点。

后者直接根据递推: dp[ i ] [ j ] = min(dp[i ] [ j ], dp[ i ] [ k ]+dp[ k+1] [ j ]);来做。

两者差异很大，但是感觉后者的不会错，实际上却是有问题的。

因为在更新的过程中，用到的  dp[ k+1] [ j ]，可能是没有被更新的。

当dp[ k+1] [ j ]被更新的时候，包含他的所有的dp[ ] [ ] 都没有被更新。

所有要长度开始枚举。

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

int ans[102];
int dp[102][102];
char a[103],b[103];
int main()
{
    int i,j,n,len,k;
    while(scanf("%s%s",a,b)>0)
    {
        n=strlen(a);
        memset(dp,0,sizeof(dp));
        memset(ans,0,sizeof(ans));

        for(i=0;i<n;i++)
            dp[i][i]=1;
        for(len=2;len<=n;len++)
        {
            for(j=len-1;j<n;j++)
            {
                i=j-len+1;
                dp[i][j]=dp[i+1][j]+1;
                for(k=i+1;k<=j;k++)
                {
                    if(b[i]==b[k])
                    {
                        if(k==j)
                            dp[i][j]=min(dp[i][j],dp[i+1][k]);
                        else
                            dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]);
                    }
                }
            }
        }
        for(i=0;i<n;i++)
            ans[i]=dp[0][i];
        for(i=0;i<n;i++)
        {
            if(a[i]==b[i])
                ans[i]=ans[i-1];
            else
            {
                for(j=0;j<=i;j++)
                    ans[i]=min(ans[i],ans[j]+dp[j+1][i]);
            }
        }
        printf("%d
",ans[n-1]);
    }
    return 0;
}