【POJ1159】Palindrome (dynamic programming)

Title: 【POJ1159】Palindrome

This question is a relatively watery DP question.

We can use f[i][j] to denote the minimum number of letters needed to turn the interval [i..j] into a palindrome, so we can easily get the state transition equation:

At the time of this s [i] == s [j], f [i] [j] = min {f [i + 1] [j-1], f [i + 1] [j] + 1, f [i ] [j-1] +1}; When this s [i]! = S [j], f [i] [j] = min {f [i + 1] [j] + 1, f [i] [ j-1] +1}

This gives the code:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#define min(x,y) (x<y?x:y)
#define N 5000
using namespace std;
int n,f[N+5][N+5];
string s;
intmain()
{
	scanf("%d",&n),cin>>s;
	for(int i=n-1;i>=0;i--) for(int j=i+1;j<n;j++) if(s[i]==s[j]) f[i][j]=min(f[i+1][j-1],min(f[i+1][j],f[i][j-1])+1);else f[i][j]=min(f[i+1][j],f[i][j-1])+1;//DP过程
	printf("%d",f[0][n-1]);
	return 0;
}

However, such code will exceed memory, what should we do?

We can change the type of the f[][] array to short int, so that there is enough memory (of course, we can also use the array for rolling storage):

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#define min(x,y) (x<y?x:y)
#define N 5000
using namespace std;
int n;
short f[N+5][N+5];
string s;
intmain()
{
	scanf("%d",&n),cin>>s;
	for(int i=n-1;i>=0;i--) for(int j=i+1;j<n;j++) if(s[i]==s[j]) f[i][j]=min(f[i+1][j-1],min(f[i+1][j],f[i][j-1])+1);else f[i][j]=min(f[i+1][j],f[i][j-1])+1;
	printf("%d",f[0][n-1]);
	return 0;
}

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