Poj1159: palindrome [DP]

Question: How many characters can be added to a string to make it a return string?

Idea: the obvious equation is: DP [I] [J] = min {DP [I + 1] [J], DP [I] [J-1], DP [I + 1] [J-1] (STR [I] = STR [J )}

DP [I] [J] indicates that the characters from I to J are constructed as the characters with the least number of characters added to the input string, but discuss says that this will exceed space + timeout

After observing this equation, the three sub-States are used each time, so the form of the equation is changed as appropriate:

DP [I] [J] indicates the minimum number of characters to be added to a string consisting of substrings whose start length is J, the equation becomes DP [I] [J] = min {DP [I + 1] [J-1], DP [I] [J-1], DP [I + 1] [J-2] (STR [I] = STR [J )}, recursive only need to store the length of J-1 and J-2 data can release the length of J data, using the pointer to change the order of the next three Arrays can be. As a result, the space problem was solved smoothly.

// Poj1159

# Include <cstdio>

# Include <string. h>

# Include <iostream>

Using namespace STD;

Const int maxn = 5009;

Int min (int A, int B)

{

If (A> B) return B; else return;

}

Int main ()

{

Int N;

Char STR [maxn];

Scanf ("% d % s", & N, STR );

Int A [maxn] = {0}, B [maxn] = {0}, C [maxn] = {0 };

Int * ans = A, * f = B, * FF = C, * temp;

For (int K = 1; k <= N; k ++)

{

For (INT I = 1; I <= n-k + 1; I ++)

{

Ans [I] = min (F [I], F [I + 1]) + 1;

If (STR [I-1] = STR [I + K-2]) ans [I] = min (ANS [I], FF [I + 1]);

}

Temp = ff; FF = f; F = ans; ans = temp;

}

Printf ("% d \ n", F [1]);

Return 0;

}

Poj1159: palindrome [DP]