标签：const while span clu family min typedef target cas
多组询问，每组给出一个字符串，求该字符串最长回文串的长度
数据范围支持$O(nlog n)$
二分+hash
回文串分奇数串和偶数串。对于奇数串，我们枚举它的中点，二分一下这个中点可以向两边扩展多远的距离；对于偶数串，我们枚举它中间两个点靠左的点，同样二分可以扩展的距离，这样时间复杂度就是$O(nlog n)$的了
说起来容易，写起来不是很容易
每次跑一遍manacher就好了
说起来容易，写起来也很容易
#include<algorithm> #include<cstring> #include<iostream> #include<cstdio> typedef unsigned long long ull; using std::string; using std::cin; using std::max; using std::min; const int N=1e6+15; int cas,ans; ull p[N],f[N],d[N]; string ch; int main() { p[0]=1; for (int i=1;i<=N;i++) p[i]=p[i-1]*113; while (cin>>ch) { if (ch=="END") break; printf("Case %d: ",++cas); memset(f,0,sizeof(f)); memset(d,0,sizeof(d)); int n=ch.size(); for (int i=0;i<n;i++) { f[i+1]=f[i]*113+ch[i]-‘a‘+1; } for (int i=n;i>=1;i--) { d[i]=d[i+1]*113+ch[i-1]-‘a‘+1; } ans=0; for (int i=1;i<=n;i++) { int l=1,r=min(i,n-i+1); while (l<r) { int mid=l+r>>1; int L1=i-mid+1,R1=i; int L2=i,R2=i+mid-1; int tmp1=f[R1]-f[L1-1]*p[R1-L1+1],tmp2=d[L2]-d[R2+1]*p[R2-L2+1]; if (tmp1==tmp2) l=mid+1; else r=mid; } int L1=i-l+1,R1=i; int L2=i,R2=i+l-1; int tmp1=f[R1]-f[L1-1]*p[R1-L1+1],tmp2=d[L2]-d[R2+1]*p[R2-L2+1]; if (tmp1!=tmp2) l--; ans=max(ans,2*l-1); l=1,r=min(i,n-i); while (l<r) { int mid=l+r>>1; int L1=i-mid+1,R1=i; int L2=i+1,R2=i+1+mid-1; int tmp1=f[R1]-f[L1-1]*p[R1-L1+1],tmp2=d[L2]-d[R2+1]*p[R2-L2+1]; if (tmp1==tmp2) l=mid+1; else r=mid; } L1=i-l+1,R1=i; L2=i+1,R2=i+1+l-1; tmp1=f[R1]-f[L1-1]*p[R1-L1+1],tmp2=d[L2]-d[R2+1]*p[R2-L2+1]; if (tmp1!=tmp2) l--; ans=max(ans,2*l); } printf("%d\n",ans); } return 0; }
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<iostream>
const int N=2e6+15;
using std::string;
using std::cin;
using std::min;
using std::max;
string ch;
char s[N];
int hw[N];
int main()
{
int cas=0;
while (cin>>ch)
{
if (ch=="END") return 0;
printf("Case %d: ",++cas);
int n=ch.size();
memset(hw,0,sizeof(hw));
s[0]=s[1]=‘#‘;
for (int i=1;i<=n;i++)
{
s[i<<1]=ch[i-1];
s[i<<1|1]=‘#‘;
}
n=n*2+2;
s[n]=0;
int mx=0,mid;
for (int i=1;i<n;i++)
{
if (i<mx) hw[i]=min(mid+hw[mid]-i,hw[(mid<<1)-i]);
else hw[i]=1;
for (;s[i+hw[i]]==s[i-hw[i]];hw[i]++);
if (hw[i]+i>mx)
{
mx=hw[i]+i;
mid=i;
}
}
int ans=1;
for (int i=1;i<n;i++) ans=max(ans,hw[i]);
printf("%d\n",ans-1);
}
return 0;
}
[poj3974] Palindrome 解题报告 （hash\manacher）
标签：const while span clu family min typedef target cas
原文地址：https://www.cnblogs.com/xxzh/p/9683597.html
