HDU 2476 String painter 区间dp

http://acm.hdu.edu.cn/showproblem.php?pid=2476

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5621    Accepted Submission(s): 2674

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

 题意：给你两个字符串，让你把第一个串变成第二个串，改变规则为：每次可以挑选一个区间，然后可以将区间里个每个字符都改变成同一个任意字符。问你至少要多少次，才能将第一个串变成第二个串？

思路：区间dp
因为第一个串的字母若想留下来，那么此字母则将整段字符串分为三部分，中间的一个字母留下来，两边的区间各自改变，并且不能对中间的字母进行任何操作。所以第一个字符串先不考虑，最后在统计相同的字母。

然后就变成了涂色问题，https://cn.vjudge.net/problem/LightOJ-1422，跟这道题一模一样。

代码：

#include <cstring>
#include <stdio.h>
#include <cctype>
#include <stdlib.h>
#include <string>
#include <map>
#include <iostream>
#include <sstream>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#include<list>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define inf 0x3f3f3f3f
typedef long long LL;
const int N=1e2+20;
const double pi=acos(-1);
int n,t;
char a[N],b[N];
int dp[N][N];
int main()
{
    while(~scanf("%s%s",a+1,b+1))
    {
        n=strlen(a+1);
        mem(dp,0);
        for(int j=1; j<=n; j++) //dp[i][j]代表区间【i,j】最小需要 涂 的次数。
            for(int i=1; i<=j; i++)
            {
                dp[i][j]=dp[i][j-1]+1; //将第j位涂色
                for(int k=i; k<j; k++) //枚举k，表示第k位和第j位可以在一次刷色刷成，
                    if(b[k]==b[j]) 
                        dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j-1]);//[i,k]+[k+1,j-1]+0(第j位不用刷色了，因为在刷第k位时就在j位也一块刷色了)，然后[k+1,j-1]单独拿出刷色
            }
        for(int k=1; k<=n; k++) //dp[i][j]代表区间【i,j】最小需要 更改 的次数。
            if(a[k]==b[k])//枚举可能不用刷色的位置
                for(int j=k; j<=n; j++)
                    dp[1][j]=min(dp[1][j],dp[1][k-1]+dp[k+1][j]);//此字母将[1,j]分成三段
        printf("%d\n",dp[1][n]);
    }
}