NYOJ-308-Substring（第四届河南省程序设计大赛F题（简单状压dp）），

NYOJ-308-Substring（第四届河南省程序设计大赛F题（简单状压dp）），

Substring

时间限制：1000 ms  |  内存限制：65535 KB 难度：1

描述

You are given a string input. You are to find the longest substring of input such that the reversal of the substring is also a substring of input. In case of a tie, return the string that occurs earliest in input. 

Note well: The substring and its reversal may overlap partially or completely. The entire original string is itself a valid substring . The best we can do is find a one character substring, so we implement the tie-breaker rule of taking the earliest one first.

输入

The first line of input gives a single integer, 1 ≤ N ≤ 10, the number of test cases. Then follow, for each test case, a line containing between 1 and 50 characters, inclusive. Each character of input will be an uppercase letter ('A'-'Z').

输出

Output for each test case the longest substring of input such that the reversal of the substring is also a substring of input

样例输入

3                   
ABCABA
XYZ
XCVCX

样例输出

ABA
X
XCVCX

题解：

题目大意：注意，这道题不是最长回文字符串。而是从原串中找一个子串，反转后使其与原串匹配最大长度的字串。

题解 ：只需将原串整个反转一下求最长公共字串。简单的dp。

下面附上代码：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int main()
{
    int n, i, j, k, s, a[2][55], len, maxx;
    char ch[55];
    scanf("%d\n", &n);
    while(n--)
    {
        gets(ch);
        memset(a, 0, sizeof(a));
        len=strlen(ch);
        for(i=len-1, maxx=0; i >= 0; i--)
            for(j=0, k=i%2; j < len; j++)
                if(ch[i] == ch[j])
                {
                    a[k][j+1]=a[!k][j]+1;//状态压缩
                    if(a[k][j+1]>=maxx)
                    {
                        maxx=a[k][j+1];
                        s=i;
                    }
                }
                else a[k][j+1]=0;
        for(i=0; i<maxx; i++)
            printf("%c",ch[i+s]);
        cout<<endl;
    }
}