HDU 3831 DICS

Problem Description
This time, we are involved in a more complicated one: given two strings SOURCE and DESTINATION , we have 4 operations
Delete: remove the ith character.
Insert: in pos i(the ith character in the first string),you can insert a character, any one as you like.
Change: in pos i,you can change that original character to any other character(of course you won't change it to itself).
Suffix change: which means you can change all characters to an identical character, from the ith pos to the end. For instance: abcdefg, you can make this operation in character b,maybe you can use d and the result is adddddd, or if you use e the result is aeeeeee or if you use this operation in character c with character f,the result is abfffff.
With the help of Suffix change, sometimes the edit distance can be greatly shortened. See the following example

abcdefg
ccccccg

You can first replace every character from the beginning to the end with c,and the first string will be ccccccc,then change the last character c to g. In such way,we can see the edit distance is 2. 

Input
Input consists of several test cases, each case contains two line,the first line will show you the SOURCE string, the second line provide the DESTINATION string. You should write a program to caculate the minimum number of operations(delete, insert, changes, suffix change as I mentioned before) of transforming the SOURCE to DESTINATION. You can assume none of string in any case will be longer than "500". Input ends with a line consists of a '#'.
PAY ATTENTION:Both strings consist of either lower or upper letters,in other word in each position there is 52 possibilities.There will be at most 15 test cases.

Output
One line for each case,the edit distance. Output nothing for the "#".

Sample Input
abcdefg
aaaaaaa

#

Sample Output

1

#include <stdio.h>#include <stdlib.h>#include <math.h>#include <string.h>#define MAXN 505char sor[MAXN], des[MAXN];int f[MAXN][MAXN][55];int leni, lenj;int imin(int a, int b){return a < b ? a : b;}int dfs(int i, int j, int k){int p;if (i >= leni) return lenj - j;if (j >= lenj) return leni - i;if (f[i][j][k] >= 0) return f[i][j][k];f[i][j][k] = 0x7fffffff;f[i + 1][j + 1][des[j + 1]] = dfs(i + 1, j + 1, des[j + 1]);f[i + 1][j + 1][52] = dfs(i + 1, j + 1, 52);f[i + 1][j + 1][k] = dfs(i + 1, j + 1, k);f[i + 1][j][des[j]] = dfs(i + 1, j, des[j]);f[i + 1][j][k] = dfs(i + 1, j, k);f[i][j + 1][des[j + 1]] = dfs(i, j + 1, des[j + 1]);f[i][j + 1][k] = dfs(i, j + 1, k);//changeif (k == 52){p = imin(f[i + 1][j + 1][52] + 1, f[i + 1][j + 1][des[j + 1]] + 1 + 1);if (sor[i] == des[j]) p--;f[i][j][k] = imin(f[i][j][k], p);}else{p = imin(f[i + 1][j + 1][k] + 1, f[i + 1][j + 1][des[j + 1]] + 1 + 1 - (k == des[j + 1]));if (k == des[j]) p--;f[i][j][k] = imin(f[i][j][k], p);}//delf[i][j][k] = imin(f[i][j][k], f[i + 1][j][des[j]] + 1 + 1 - (k == des[j]));f[i][j][k] = imin(f[i][j][k], f[i + 1][j][k] + 1);//insertf[i][j][k] = imin(f[i][j][k], f[i][j + 1][des[j + 1]] + 1 + 1 - (k == des[j + 1]));f[i][j][k] = imin(f[i][j][k], f[i][j + 1][k] + 1);return f[i][j][k];}int main(){int i, j, k, p;//freopen("DICS.in","r",stdin);//freopen("out.txt","w",stdout);while(scanf("%s", sor), sor[0] != '#'){scanf("%s", des);leni = strlen(sor);lenj = strlen(des);memset(f, -1, sizeof(f));for (i = 0; i < leni; i++){if (sor[i] >= 'a' && sor[i] <= 'z') sor[i] = sor[i] - 'a' + 26;else sor[i] = sor[i] - 'A';}for (i = 0; i < lenj; i++){if (des[i] >= 'a' && des[i] <= 'z') des[i] = des[i] - 'a' + 26;else des[i] = des[i] - 'A';}p = imin(dfs(0, 0, 52), dfs(0, 0, des[0]) + 1);printf("%d\n", p);}//    system("pause");    return 0;}/***************其实状态和方程都不难想，就是写起来有点乱...思路不清晰吧...f[i][j][k] 原串后i个字符，与目标后j个字符做匹配，在这之前最近一次发生的置后缀操作为“置k”，k=52时表示不置后缀changef[i][j][k] = f[i + 1][j + 1][k1] (sor[i] == des[j])             f[i + 1][j + 1][k1] + 1注意，若k != 52 则要看 k == des[j]      若k1 != k， 说明发生了置后缀操作 再 + 1      （下同，不再赘述）delf[i][j][k] = f[i + 1][j][k1] + 1insertf[i][j][k] = f[i][j + 1][k1] + 1实际k1 只可能取 k, des[j](或des[j + 1])，取其他不是添乱么...用记忆化搜索，想减少不必要的状态，但没写好，跑起来极慢...场上脑残，把样例看错了，一直以为自己错了，在那里调啊调, 后来自己给了组数据，答案算错了，又在那里调啊调...结束后，跑了遍数据竟然A了...这世上，什么事都有= =***************/