Hdu 2476 String painter【思维+区间Dp】好题~

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4208    Accepted Submission(s): 1960

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

zzzzzfzzzzzabcdefedcbaababababababcdcdcdcdcdcd

 

Sample Output

67

题目大意：

给你两个字符串，A和B，每一次我们可以使用一种颜色（用字母代替）刷任意位子的连续子序列，问将字符串A变成字符串B最少需要刷多少次。

思路：

1、考虑到要求最小步数，考虑dp.观察到数据范围，考虑区间dp.

一开始设定dp【i】【j】表示使得区间【i，j】的A串变成B串的最小步数转移和结果相对很难得到。写了半天样例也是搞不过去。

后来反应过来，我们不妨设定dp【i】【j】表示将一个空串区间【i，j】刷成B字符串的最小步数。

那么对应我们再处理哪里进行对A串的保留，哪里进行重新刷就行了。

整体思路确定如下：

①设定dp【i】【j】表示将一个空串区间【i，j】刷成B字符串的最小步数。

②设定F【i】表示将A串将区间【1，i】刷成B串需要的最小步数。

2、那么我们首先搞定区间dp的状态转移。

不难推出有四种转移方式：

注意区间合并操作之类的细节即可。

3、那么接下来搞定F【i】的状态转移：

①如果当前A【i】==B【i】.那么这一个位子是不需要重新刷的，那么对应有：f【i】=min（f【i】，f【i-1】）；

②如果当前A【i】！=B【i】.那么我们枚举一个位子作为起点刷一个区间【j，i】即可，那么对应有：f【i】=min（f【i】，f【j】+dp【j+1】【i】）；

Ac代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;char a[150];char b[150];int dp[150][150];int f[150];int judge(int i,int j){    if(b[i]!=b[j])return 1;    else return 0;}int main(){    while(~scanf("%s%s",a+1,b+1))    {        int n=strlen(a+1);        memset(dp,0,sizeof(dp));        memset(f,0,sizeof(f));        for(int len=0;len<=n;len++)        {            for(int i=1;i<=n;i++)            {                int j=i+len;                if(len==0)                {                    dp[i][j]=1;                }                else if(j<=n)                {                    dp[i][j]=0x3f3f3f3f;                    if(i+1<=n)dp[i][j]=min(dp[i][j],dp[i+1][j]+judge(i,i+1));                    if(i+1<=n)dp[i][j]=min(dp[i][j],dp[i+1][j]+judge(i,j));                    if(j-1>=1)dp[i][j]=min(dp[i][j],dp[i][j-1]+judge(j-1,j));                    if(j-1>=1)dp[i][j]=min(dp[i][j],dp[i][j-1]+judge(i,j));                    for(int k=i;k<j;k++)                    {                        dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);                    }                }            }        }        for(int i=1;i<=n;i++)        {            f[i]=dp[1][i];            if(a[i]==b[i])f[i]=min(f[i-1],f[i]);            else            {                for(int j=0;j<i;j++)                {                    f[i]=min(f[i],f[j]+dp[j+1][i]);                }            }        }        printf("%d\n",f[n]);    }}