【回文串】HDU 3294 Girls' research manacher算法

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O(n) 复杂度

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#define cler(arr, val) memset(arr, val, sizeof(arr))
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define IN freopen ("in.txt" , "r" , stdin);
#define OUT freopen ("out.txt" , "w" , stdout);
typedef long long LL;
const int MAXN = 510011;
const int MAXM = 6000010;
const int INF = 0x3f3f3f3f;
const LL mod = 2147483647;
const double eps= 1e-8;
const double pi=acos(-1.0);
#define lson l,m, rt<<1
#define rson m+1,r,rt<<1|1
char s[201000];
char c[410200],in;
int p[410002];
int len;
void solve()
{
c[0]='@';
for(int i=0;i<len;i++)
c[i*2+1]='#',c[i*2+2]=s[i]=(s[i]-in+26)%26+'a';
len=2*len+2;
c[len-1]='#';
c[len]=0;
int mx=0,id=0;
for(int i=1;i<len;i++)
{
if(mx>i)
p[i]=min(p[2*id-i],p[id]+id-i);
else p[i]=1;
while(c[i+p[i]]==c[i-p[i]])
p[i]++;
if(i+p[i]>mx)
mx=i+p[i],id=i;
}
int ans=0;
for(int i=1;i<len;i++)
if(p[ans]<p[i]) ans=i;
int len=p[ans]-1;
if(len==1) puts("No solution!");
else
{
int l=(ans+1)/2-1-len/2;
printf("%d %d\n",l,l+len-1);
for(int i=l;i<l+len;i++)
printf("%c",s[i]);
puts("");
}
}
int main()
{
while(scanf("%c %s",&in,s)!=EOF)
{
getchar();
len=strlen(s);
solve();
}
return 0;
}