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
求最长回文串,有很多经典的算法
算法一:中心扩展法,比较好理解的算法之一,就是以s串的字符为遍历对象,结果回文串相对于每个字符只有两种可能,一种是奇对称,一种偶对称,所以再分两种情况,比较两种情况下的最大回文串的长度。
#include <iostream>
#include <string>
using namespace std;
string s;
int result=0,len;
void expandaroundcenter(int left,int right){
while(left>=0&&right<len&&s[left]==s[right]){
left--;right++;
}
result=max(right-left-1,result);
}
int main(){
getline(cin,s);
len=s.length();
for(int i=0;i<len;i++){
expandaroundcenter(i,i);
expandaroundcenter(i,i+1);
}
printf("%d",result);
}
算法二:Manancher算法
貌似有个测试点用了‘#’,emmmm所以取特殊字符的时候要留意
#include <iostream>
#include <string>
#include <vector>
using namespace std;
string s,s2,temp;
int result=0,len,len2;
int main(){
getline(cin,s);
len=s.length();
s2+='@';
for(int i=0;i<len;i++)
s2=s2+'!'+s[i];
s2+="!$0";
len2=s2.length();
std::vector<int> Len(len2,0);
int mx=0,po=0;
for(int i=0;i<len2;i++){
if(mx>i)
Len[i]=min(mx-i,Len[2*po-i]);
else
Len[i]=1;
while(s2[i-Len[i]]==s2[i+Len[i]])
Len[i]++;
if(Len[i]+i>mx){
mx=Len[i]+i;
po=i;
}
if(Len[i]>result){
result=Len[i];
temp=s.substr(i-Len[i]+1,2*Len[i]-1);
}
}
for(int i=0;i<temp.length();i++)
if(temp[i]=='!')
result--;
printf("%d",result-1);
}
算法三:动态规划
#include <iostream>
#include <string>
using namespace std;
string s;
bool dp[1010][1010];
int len,result;
int main(){
getline(cin,s);
len=s.length();
fill(dp[0],dp[0]+1010*1010,false);
for(int i=0;i<s.length();i++){
dp[i][i]=true;
if(i!=len-1&&s[i]==s[i+1]){
dp[i][i+1]=true;
result=2;
}
}
for(int k=3;k<=len;++k)
for(int i=0;i+k-1<len;i++)
if(s[i]==s[i+k-1]&&dp[i+1][i+k-2]==true){
dp[i][i+k-1]=true;
result=k;
}
printf("%d",result);
}