HDU 2476 String painter（区间DP）

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题意：给你一串字符串，一次可以修改一个或者多个连续的字符串，问最少操作多少次使得原串变成目标串

思路：一开始完全不知道怎么DP...首先我们假设一个空串，先求空串变成目标串的最少次数，令dp[i][j]为i到j的操作次数，那么显然有dp[i][j]=dp[i+1][j]+1，如果目标串有s[i]==s[k]，那么dp[i][j]=min(dp[i][j]，dp[i+1][k]+dp[k+1][j])

#include <cstdio>#include <queue>#include <cstring>#include <iostream>#include <cstdlib>#include <algorithm>#include <vector>#include <map>#include <string>#include <set>#include <ctime>#include <cmath>#include <cctype>using namespace std;#define maxn 100000#define LL long longint cas=1,T;int dp[105][105],a[105];char s[105],t[105];int main(){while (scanf("%s",s+1)!=EOF){scanf("%s",t+1);s[0]=t[0]=2;int len = strlen(s)-1;memset(dp,0,sizeof(dp));memset(a,0,sizeof(a));for (int i = 1;i<=len;i++)dp[i][i]=1;for (int i = len-1;i>=1;i--){for (int j = i+1;j<=len;j++){dp[i][j]=dp[i+1][j]+1;for (int k = i+1;k<=j;k++)if (t[i]==t[k])dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]);}}for (int i = 1;i<=len;i++){a[i]=dp[1][i];if (s[i]==t[i]){if (i==1)a[i]=0;elsea[i]=a[i-1];}elsefor (int j = 1;j<i;j++)a[i]=min(a[i],a[j]+dp[j+1][i]);}printf("%d\n",a[len]);}//freopen("in","r",stdin);//scanf("%d",&T);//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);return 0;}

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

zzzzzfzzzzzabcdefedcbaababababababcdcdcdcdcdcd

Sample Output

0 0