HDU

Problem Describe

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

题意 : 每次能将一个区间里面的字母改成一个同一个字母，问最少需要多少次将str1串变成str2串

思路 : 这个题感觉很怪 ，明显的区间dp ，可是偏偏无从下手 ， 它每次dp的时候str1 串 和 str2串匹配的时候总是感觉很难搞 ， 用平常的区间dp写了一下 WA了 ， 于是菜的我去搜了题解 发现区间 dp 还可以这样搞 ，神奇 呐~~

AC code :

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

using namespace std;

const int maxn = 110 ;

int dp[maxn][maxn] ,len ,ans[maxn] ;
char str1[maxn] ,str2[maxn] ;

int main() {
	while(~scanf("%s %s",str1 + 1 ,str2 + 1 )) {
		len = strlen(str1 + 1) ;
		memset(dp ,0 , sizeof(dp) ) ;
		memset(ans ,0 ,sizeof(ans) ) ;
		for (int i = 1 ; i <= len ; i ++) {
			for (int j = i ; j <= len ; j ++ ) {
				dp[i][j] = j - i + 1 ;
			}
		}
		for (int l = 2 ; l <= len ; l ++ ) {
			for (int i = 1 ; i <= len - l + 1 ; i ++) {
				int j = i + l - 1;
				if ( str2[i] == str2[j] ) dp[i][j] = dp[i][j-1];
				else dp[i][j] = dp[i][j-1] + 1;
				for (int k = i ; k <= j - 1 ; k ++ ) {
					dp[i][j] = min(dp[i][j] ,dp[i][k] + dp[k+1][j] ) ;
				}
			}
		} 
		for (int i = 1 ; i <= len ; i ++ ) ans[i] = dp[1][i] ;
		for (int i = 1 ; i <= len ; i ++ ) {
			if ( str1[i] == str2[i] ) {
				ans[i] = ans[i-1];
			}
			for (int j = 1 ; j <= i - 1 ; j ++ ) {
				ans[i] = min(ans[i] ,ans[j] + dp[j+1][i] );
			}
		}
		printf("%d
",ans[len]);
	}
	return 0;
}