标签：must cat ati contains const 基础 key you div
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
思路
求一个字符串的最长回文子串。
DP的思想，复杂度O(n^2)。
1.如果一个从i到j的字符串str（i,j）的子串str(i+1,j-1)为回文串，那么在str[i] == str[j]的情况下，字符串str(i,j)也是回文串。
2.每一个字符本身就是一个回文串。所以在每一个字符的基础上，根据1的条件来确定更长的回文串。
3.用一个bool数组isSym[1001][1001]来列举所有的情况，isSym[i][j]表示起始位置为i、终止位置为j的字符串是否是回文串。
4.检查是否有N个长度的回文串，更新最大长度maxlength。（1 <=N <= str.length()）。
代码
#include<iostream>
#include<vector>
using namespace std;
vector<vector<bool>> isSym(1001,vector<bool>(1001,false));
int main()
{
string s;
getline(cin,s);
int maxlength = 1;
const int Length = s.size();
for(int i = 0;i < Length;i++)
{
isSym[i][i] = true;
if(i < Length - 1 && s[i] == s[i + 1])
{
isSym[i][i+1] = true;
maxlength = 2;
}
}
for(int len = 3;len <= Length;len++)
{
for(int i = 0;i <= Length - len;i++)
{
int j = i + len - 1;
if(isSym[i+1][j-1] && s[i] == s[j])
{
isSym[i][j] = true;
maxlength = len;
}
}
}
cout << maxlength << endl;
}
PAT1040:Longest Symmetric String
标签：must cat ati contains const 基础 key you div
原文地址：http://www.cnblogs.com/0kk470/p/7769065.html
