HDU 2476 String painter

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3639    Accepted Submission(s): 1697

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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题目大意：给定两个等长度的字符串，有一种刷新字符串的方法，它能够将一段字符串刷成同一个字符(任意字符)。现在要你使用这种方法，使得第一个字符串被刷成第二个字符串，问你最少需要刷多少次？

 

解题思路：显然这是一道区间DP的题。设dp[i][j]表示区间[i,j]内最少需要刷多少次。直接确定状态转移方程不太好确定，所以我们需要考虑直接将一个空串刷成第二个字符串，然后再与第一个字符串去比较。这样，如果每个字符都是单独刷新，则dp[i][j] = dp[i+1][j]+1，如果在区间[i+1,j]之间有字符与t[i]相同,则可以将区间分为两个区间，分别为[i+1,k]和[k+1,j]，考虑一起刷新。详见代码。

 

附上AC代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 105;
char s[maxn], t[maxn];
int dp[maxn][maxn];

int main(){
    while (~scanf("%s%s", s, t)){
        memset(dp, 0, sizeof(dp));
        int len = strlen(s);
        for (int j=0; j<len; ++j)
            for (int i=j; i>=0; --i){
                dp[i][j] = dp[i+1][j]+1;        // 每一个都单独刷
                for (int k=i+1; k<=j; ++k)
                    if (t[i] == t[k])            // 区间内有相同颜色,考虑一起刷
                        dp[i][j] = min(dp[i][j], dp[i+1][k]+dp[k+1][j]);
            }
        for (int i=0; i<len; ++i){
            if (s[i] == t[i]){    // 对应位置相同,可以不刷
                if (i)
                    dp[0][i] = dp[0][i-1];
                else
                    dp[0][i] = 0;
            }
            else
                for (int j=0; j<i; ++j)        // 寻找当前区间的最优解
                    dp[0][i] = min(dp[0][i], dp[0][j]+dp[j+1][i]);
        }
        printf("%d
", dp[0][len-1]);
    }
    return 0;
}