题意

String painter

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6669 Accepted Submission(s): 3230

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

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分析

参照AndyQsmart的题解。

读完题首先会想到的自然是用区间dp，但是列出来发现，没办法区间合并。因为一旦需要考虑对某一段成段染色的话，在区间合并的时候，就无法考虑转移过程中起始串的变化了。

既然这样，就不考虑成段染色造成的影响了，就当起始串和目标串处处不想等。

那么考虑区间[i, i+len]，

自然遍历子区间[i, j]，

如果[i, j]和[j+1, i+len]需要合并的话，

如果考虑成段染色的话，只有str2[i] == str2[j+1]时，考虑成段染色[i, j+1]，但是[i, j+1]的父区间又有可能会成段然和str2[i]一样的颜色，所以不能直接将区间缩短成[i+1, j]和[j+2, i+len]，所以可以考虑这一步的效果只相当于染str2[j+1]的时候，可以少染一个str2[i]。那么区间就变成[i+1, j]和[j+1, i+len]，这样父区间中可能再次出现一个i`，和j+1产生成段染色，即

p[i][i+len] = min(p[i][i+len], p[i+1][j]+p[j+1][i+len]);

然后就是考虑使用p来计算ans[i],表示前i个字符从起始串到目标串的步数。

ans[0]自然好考虑，只需要判断一下str1[0]和str2[0]。

对于ans[i]，

如果str1[i] == str2[i]，自然就可以退化成ans[i-1]。

其它情况，自然是遍历子区间ans[j]和p[j+1][i]进行合并。

时间复杂度\(O(n^3)\)

代码

#include<iostream>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;
    rg char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-') w=-1;
        ch=getchar();
    }
    while(isdigit(ch))
        data=data*10+ch-'0',ch=getchar();
    return data*w;
}
template<class T>il T read(rg T&x){
    return x=read<T>();
}
typedef long long ll;

char str1[105],str2[105];
int n,p[105][105],ans[105];
void work(){
    for(int i=0;i<n;++i) p[i][i]=1;
    for(int len=1;len<n;++len)
        for(int i=0;i<n&&i+len<n;++i){
            p[i][i+len]=p[i+1][i+len]+1;
            for(int j=i;j<i+len;++j)if(str2[i]==str2[j+1])
                p[i][i+len]=std::min(p[i][i+len],p[i+1][j]+p[j+1][i+len]);
        }
    ans[0]=str1[0]==str2[0]?0:1;
    for(int i=1;i<n;++i){
        ans[i]=str1[i]==str2[i]?ans[i-1]:p[0][i];
        for(int j=0;j<i;++j)
            ans[i]=std::min(ans[i],ans[j]+p[j+1][i]);
    }
    printf("%d\n",ans[n-1]);
}
int main(){
//  freopen(".in","r",stdin),freopen(".out","w",stdout);
    while(~scanf("%s%s",str1,str2)){
        n=strlen(str1);
        work();
    }
    return 0;
}

巴特西

HDU2476 String painter

题意

String painter

分析

代码

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