1040. Longest Symmetric String (25)

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

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题目大意：

给定一个字符串，你应该输出最长对称字符串的长度。例如，给定"Is PAT&TAP symmetric?"，因此你必须输出11.
输入规格：
数个输入文件包含一个测试用例，该测试用例的长度不超过1000.
输出规范：
对于每个测试用例，只需在一行中打印最大的长度。

代码：

解法一：

#include<string>
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
int main()
{
    int i,j,n,m,k,t,Max=1;
    string s;
    getline(cin,s);
    for(i=0;i<s.size();i++)
    {
        if(s.size()-i<=Max)
            break;
        for(j=i;j<s.size();j++)
        {
            string str1,str2;
            str1=s.substr(i,j+1);
            str2=str1;
            reverse(str1.begin(),str1.end());
            if(str1==str2)
            {
                if(str1.size()>Max)
                {
                    Max=str1.size();
                }
            }

        }
    }
    printf("%d\n",Max);
    return 0;
}

解法二：

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int dp[1010][1010];
int main()
{
    string str;
    int i,j,m,k,t;
    getline(cin,str);
    fill(dp[0],dp[0]+1010*1010,1);
    int n=str.length();
    //printf("%d\n",n);
    int Max=1;
    for(i=0;i<n-1;i++)
    {
        if(str[i]==str[i+1])
            {
                dp[i][i+1]=1;
                Max=2;
            }
            else
            {
               dp[i][i+1]=0;
            }
    }
    for(i=2;i<n;i++)
    {
        for(j=0;j+i<n;j++)
        {
                if(str[j]==str[j+i]&&dp[j+1][j+i-1]!=0)
                {
                    //printf("%c %c\n",str[j],str[j+i]);
                    dp[j][j+i]=1;
                    Max=i+1;
                }
                else
                {
                    dp[j][j+i]=0;
                }
        }
    }
    printf("%d\n",Max);
    return 0;
}