PAT 1040. Longest Symmetric String (25)

http://pat.zju.edu.cn/contests/pat-a-practise/1040

Given a string, you are supposed to output the length of the longest symmetric sub-string. For example, given "Is PAT&TAP symmetric?", the longest symmetric sub-string is "s PAT&TAP s", hence you must output 11.

Input Specification:

Each input file contains one test case which gives a non-empty string of length no more than 1000.

Output Specification:

For each test case, simply print the maximum length in a line.

Sample Input:

Is PAT&TAP symmetric?

Sample Output:

11

遍历字符串，以各个字符为对称中心的土鳖解法（这里要注意到对称序列长度有单双两种情况）：

#include 
#include 
#include 
#include 
using namespace std;
int main(){
	string s;
	getline(cin, s);
	int len = 1;
	for (int i = 0; i < s.size(); i++){
		for (int j = 0; j <= min(i, int(s.size())-1-i); j++){
			string s2 = s.substr(i-j,2*j+1);
			string s3 = s2;
			reverse(s2.begin(), s2.end());
			if (s2 == s3){
				if (s2.size() > len)
					len = s2.size();
			}
			string s4 = s.substr(i-j, 2*j+2);
			string s5 = s4;
			reverse(s4.begin(), s4.end());
			if (s4 == s5){
				if (s4.size() > len)
					len = s4.size();
			}
		}
	}
	printf("%d\n", len);
	return 0;
}

简单点的写法：

#include 
#include 
#include 
#include 
using namespace std;
int main(){
	string s;
	getline(cin, s);
	int len = 0;
	for (int i = 0; i < s.size(); i++){
		int m = 0, n = 0, len1 = 0, len2 = 0;
		while(i - n >= 0 && i + n < s.size() && s[i - n] == s[i + n]){
			len1 = 2 * n + 1;
			n++;
		}
		if (len1 > len) len = len1;
		while(i + 1 < s.size() && s[i] == s[i + 1] && i - m >= 0 && i + m + 1 < s.size() && s[i - m] == s[i + 1 + m]){
			len2 = 2 * m + 2;
			m++;
		}
		if (len2 > len) len = len2;
	}
	printf("%d\n", len);
	return 0;
}

网上看到的动态规划解法：

using namespace std;

int main()
{
	string s;
	getline(cin,s);	//读取包含空格的一行
	int max=1;
	int n=s.size();	//字符串长度
	bool flag[1005][1005];	//s[i]~s[j]是否是回文，1是，0不是

	int i,j;
	for(i=0;imax)
					max=j-i+1;
			}
		}
	}
	cout<