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[20005];
int main(){
    //freopen("D:\INPUT.txt","r",stdin);
    //scanf("%s",s);
    string s;
    getline(cin,s);

    //cout<<s<<endl;

    int i,j=0,k,count=0;
    dp[j++]=-1;
    for(i=0;i<s.length();i++){
        dp[j++]=s[i];//hash
        dp[j++]=-1;
    }

    //cout<<j<<endl;

    for(i=1;i<j;i++){//i从1开始！！
        int f=i-1,b=i+1;
        while(f>=0&&b<j&&dp[f]==dp[b]){
            f--;
            b++;
        }
        if(count<b-f-1){
            count=b-f-1;
            //cout<<count<<endl;
        }
    }
    printf("%d
",count/2);//这里可以分为中心为-1和正常数字 2种情况讨论
    return 0;
}

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=0;
    for(i=0;i<s.length();i++){
        int f=i,b=i;
        while(f>=0&&b<s.length()&&s[f]==s[b]){
            f--;
            b++;
        }
        if(count<b-f-1){
            count=b-f-1;
        }
        f=i;
        b=i+1;
        while(f>=0&&b<s.length()&&s[f]==s[b]){
            f--;
            b++;
        }
        if(count<b-f-1){
            count=b-f-1;
        }
    }
    printf("%d
",count);
    return 0;
}