[leetcode]664. Strange Printer

题目链接：https://leetcode.com/problems/strange-printer/description/

There is a strange printer with the following two special requirements:

1. The printer can only print a sequence of the same character each time.
2. At each turn, the printer can print new characters starting from and ending at any places, and will cover the original existing characters.

Given a string consists of lower English letters only, your job is to count the minimum number of turns the printer needed in order to print it.

Example 1:

Input: "aaabbb"Output: 2Explanation: Print "aaa" first and then print "bbb".

Example 2:

Input: "aba"Output: 2Explanation: Print "aaa" first and then print "b" from the second place of the string, which will cover the existing character 'a'.

Hint: Length of the given string will not exceed 100.

This problem is simiiar to #546 Remove Boxes which uses f[l][r][k] to store the maximum points of range [l, r] with k boxes equal to r. But for this problem, we can use 2D-array DP instead of 3D-array DP because the store of k is useless.

f[i][j] represents the minumum turns to print the sequence from i to j. The transition function should be:

f[i][j] = min(f[i][k] + f[k+1][j-1]) for each k where i<k<j and s[k]=s[j]

Do not forget the common transition:

f[i][j] = f[i][j-1] + 1

And the border condition:

f[i][j] = 0 where i>j

class Solution {public:    int strangePrinter(string s) {        memset(f,0, sizeof(f));        int size=s.size();        return dfs(s,0,size-1);    }private:    int f[100][100];    int dfs(string &s,int l,int r)    {        if(l>r) return 0;        if(f[l][r]) return f[l][r];        f[l][r]=dfs(s,l,r-1)+1;        for(int i=l;i<r;i++)        {            if(s[i]==s[r])            {                f[l][r]=min(f[l][r],dfs(s,l,i)+dfs(s,i+1,r-1));            }        }        return f[l][r];    }};