hdu 2476 String painter(区间dp)

                                             String painter

                                                                                               Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                                      Total Submission(s): 4108    Accepted Submission(s): 1915
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

题意：两个字符串AB A->B按照规则变  规则是将A字符串中【l r】区间的字符变成一个字符  问最小的变换次数

我们先用一个空串用区间dp按照规则去求出空串变成B需要的至少变换次数（dp【0】【len-1】）

最后我们根据AB字符串选择dp

if(a[i]==b[i])dp[0][i]=dp[0][i-1];

else dp[0][i]=min(dp[0][i],dp[0][j]+dp[j+1][i])    (j<i);

 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<cstring>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<algorithm>
#include<string.h>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=0.0000000001;
const int N=1000+10;
char a[N],b[N];
int dp[N][N];
int DP[N];
int main(){
    while(gets(a)){
        gets(b);
        int len=strlen(a);
        memset(dp,INF,sizeof(dp));
        for(int i=0;i<len;i++)dp[i][i]=1;
        for(int t=1;t<len;t++){
            for(int i=0;i+t<len;i++){
                int j=i+t;
                if(b[i]==b[j])dp[i][j]=dp[i][j-1];
                else{
                    for(int k=i;k<j;k++)
                    dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);
                }
            }
        }
        for(int i=0;i<len;i++){
            if(i==0&&a[i]==b[i])dp[0][i]=0;
            else if(a[i]==b[i])dp[0][i]=dp[0][i-1];
            for(int j=0;j<i;j++)
            dp[0][i]=min(dp[0][i],dp[0][j]+dp[j+1][i]);
        }
        cout<<dp[0][len-1]<<endl;
    }
}