HDU 2476 String Painter

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String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem 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

 

Source

2008 Asia Regional Chengdu

比较好的区间dp，可以先求空串到目标串的最少操作数

然后再对原串进行dp

#include<set>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN=110;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)
#define Down(i,r,l) for(int i=r;i>=l;i--)
int dp[MAXN][MAXN];
char str1[MAXN],str2[MAXN];
int ans[MAXN];
int main(){
    while(~scanf("%s%s",str1+1,str2+1)){
        int n=strlen(str1+1);
        memset(dp,0,sizeof(dp));
        For(i,n)
            Rep(j,i,n) dp[i][j]=j-i+1;
        Down(i,n-1,1)
            Rep(j,i+1,n){
                dp[i][j]=dp[i+1][j]+1;
                Rep(k,i+1,j)
                    if(str2[i]==str2[k])
                        dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k][j]);
            }
        For(i,n){
            ans[i]=dp[1][i];
            if(str1[i]==str2[i])  ans[i]=ans[i-1];
            For(j,i)              ans[i]=min(ans[i],ans[j]+dp[j+1][i]);
        }
        printf("%d
",ans[n]);
    }
    return 0;
}