pat1040. Longest Symmetric String (25)

1040. Longest Symmetric String (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

---

提交代码

方法一：插入无效字符，遍历一次即可。

 #include<cstdio>
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<queue>
 #include<vector>
 #include<cmath>
 #include<string>
 #include<map>
 #include<set>
 using namespace std;
 int dp[];
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     //scanf("%s",s);
     string s;
     getline(cin,s);

     //cout<<s<<endl;

     int i,j=,k,count=;
     dp[j++]=-;
     for(i=;i<s.length();i++){
         dp[j++]=s[i];//hash
         dp[j++]=-;
     }

     //cout<<j<<endl;

     for(i=;i<j;i++){//i从1开始！！
         int f=i-,b=i+;
         while(f>=&&b<j&&dp[f]==dp[b]){
             f--;
             b++;
         }
         if(count<b-f-){
             count=b-f-;
             //cout<<count<<endl;
         }
     }
     printf("%d\n",count/);//这里可以分为中心为-1和正常数字 2种情况讨论
     return ;
 }

方法二：分奇偶别讨论：

 #include<cstdio>
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<queue>
 #include<vector>
 #include<cmath>
 #include<string>
 #include<map>
 #include<set>
 using namespace std;
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     //scanf("%s",s);
     string s;
     getline(cin,s);

     //cout<<s<<endl;

     int i,j,k,count=;
     for(i=;i<s.length();i++){
         int f=i,b=i;
         while(f>=&&b<s.length()&&s[f]==s[b]){
             f--;
             b++;
         }
         if(count<b-f-){
             count=b-f-;
         }
         f=i;
         b=i+;
         while(f>=&&b<s.length()&&s[f]==s[b]){
             f--;
             b++;
         }
         if(count<b-f-){
             count=b-f-;
         }
     }
     printf("%d\n",count);
     return ;
 }