LA 4394

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

令dp[l][r] 为l到r区间从空串刷成b串所需的最少次数
则dp[l][r] = min(dp[l][k] + dp[k + 1][r])
如果 b[l] == b[r] dp[l][r] = min(dp[l][r], dp[l][r - 1])

最后设f[i]为1 到 i从a串刷到b串的最少次数(下标从1开始）
则 f[i] = min(f[j] + dp[j][i - 1])
如果a[i] == b[i]则 f[i] = min(f[i], f[i - 1])

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>

using namespace std;

#define read() freopen("sw.in", "r", stdin)

const int MAX = 105;
char a[MAX], b[MAX];
int dp[MAX][MAX];
int f[MAX];

void solve() {
        int n = strlen(a);
        for (int len = 1; len <= n; ++len) {
                for (int l = 0; l + len - 1 < n; ++l) {
                        int r = l + len - 1;
                        dp[l][r] = len;
                        if (l == r) {
                                dp[l][r] = 1;
                                continue;
                        }
                        for (int k = l; k <= r; ++k) {
                                if (k + 1 > r) continue;
                                dp[l][r] = min(dp[l][r], dp[l][k] + dp[k + 1][r]);
                        }
                        if (b[l] == b[r]) dp[l][r] = min(dp[l][r - 1], dp[l][r]);
                        //printf("l = %d r = %d %d
", l, r, dp[l][r]);
                }
        }

        for (int i = 0; i < n; ++i) {
                f[i + 1] = dp[0][i];
                if (a[i] == b[i]) f[i + 1] = min(f[i + 1], f[i]);
                for (int j = 0; j < i; ++j) {
                        if (j + 1 > i) continue;
                        f[i + 1] = min(f[i + 1], f[j + 1] + dp[j + 1][i]);
                }
        }

        //for (int i = 0; i < n; ++i) printf("%d ", f[i]);
        printf("%d
", f[n]);


}

int main()
{

    int t;
    //read();
    while (scanf("%s%s", a, b) != EOF) {
           solve();
    }
    //cout << "Hello world!" << endl;
    return 0;
}