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

方法一：插入无效字符，遍历一次即可。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<queue>
 6 #include<vector>
 7 #include<cmath>
 8 #include<string>
 9 #include<map>
10 #include<set>
11 using namespace std;
12 int dp[20005];
13 int main(){
14     //freopen("D:\\INPUT.txt","r",stdin);
15     //scanf("%s",s);
16     string s;
17     getline(cin,s);
18
19     //cout<<s<<endl;
20
21     int i,j=0,k,count=0;
22     dp[j++]=-1;
23     for(i=0;i<s.length();i++){
24         dp[j++]=s[i];//hash
25         dp[j++]=-1;
26     }
27
28     //cout<<j<<endl;
29
30     for(i=1;i<j;i++){//i从1开始！！
31         int f=i-1,b=i+1;
32         while(f>=0&&b<j&&dp[f]==dp[b]){
33             f--;
34             b++;
35         }
36         if(count<b-f-1){
37             count=b-f-1;
38             //cout<<count<<endl;
39         }
40     }
41     printf("%d\n",count/2);//这里可以分为中心为-1和正常数字 2种情况讨论
42     return 0;
43 }

方法二：分奇偶别讨论：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<queue>
 6 #include<vector>
 7 #include<cmath>
 8 #include<string>
 9 #include<map>
10 #include<set>
11 using namespace std;
12 int main(){
13     //freopen("D:\\INPUT.txt","r",stdin);
14     //scanf("%s",s);
15     string s;
16     getline(cin,s);
17
18     //cout<<s<<endl;
19
20     int i,j,k,count=0;
21     for(i=0;i<s.length();i++){
22         int f=i,b=i;
23         while(f>=0&&b<s.length()&&s[f]==s[b]){
24             f--;
25             b++;
26         }
27         if(count<b-f-1){
28             count=b-f-1;
29         }
30         f=i;
31         b=i+1;
32         while(f>=0&&b<s.length()&&s[f]==s[b]){
33             f--;
34             b++;
35         }
36         if(count<b-f-1){
37             count=b-f-1;
38         }
39     }
40     printf("%d\n",count);
41     return 0;
42 }