HDU2476

1.题目描述

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

2.题意概述：

给出两个串s1和s2，一次只能将一个区间刷一次，问最少几次能让s1=s2。

例如zzzzzfzzzzz，长度为11，我们就将下标看做0~10
先将0~10刷一次，变成aaaaaaaaaaa
1~9刷一次，abbbbbbbbba
2~8:abcccccccba
3~7:abcdddddcba
4~6:abcdeeedcab
5:abcdefedcab
这样就6次，变成了s2串了
第二个样例也一样
0
先将0~10刷一次，变成ccccccccccb
1~9刷一次，cdddddddddcb
2~8:cdcccccccdcb
3~7:cdcdddddcdcb
4~6:cdcdcccdcdcb
5:cdcdcdcdcdcb
最后竟串尾未处理的刷一次
就变成了串2cdcdcdcdcdcd
所以一共7次

3.解题思路：

集训时候直接把这题跳了，学长都写了hard两字orz，然后结束后参考了网上的题解，对于这种最优子结构问题可以考虑用区间dp做，把s1变成s2，关键是看它们相同的区间。dp问题还需要细嚼慢咽啊。

4.AC代码：

#include <stdio.h>#include <string.h>#include <algorithm>#define maxn 101using namespace std;char s1[maxn], s2[maxn];int dp[maxn][maxn], ans[maxn];int main(){while (scanf("%s%s", s1, s2) != EOF){memset(dp, 0, sizeof(dp));int len = strlen(s1);for (int i = 0; i < len; i++)for (int j = i; j >= 0; j--){dp[j][i] = dp[j + 1][i] + 1;for (int k = j + 1; k <= i; k++) // 从j到i中间的所有刷法if (s2[j] == s2[k])// 如果j和k相同，寻找j到k和k+1到i的最佳方案dp[j][i] = min(dp[j][i], dp[j + 1][k] + dp[k + 1][i]);}for (int i = 0; i < len; i++)ans[i] = dp[0][i];// ans记录0到i的最优解for (int i = 0; i < len; i++)if (s1[i] == s2[i])// 如果第i个位置相同，则它不用刷ans[i] = ans[i - 1];elsefor (int j = 0; j < i; j++)//枚举区间，取最优情况ans[i] = min(ans[i], ans[j] + dp[j + 1][i]);printf("%d\n", ans[len - 1]);}    return 0;}