BZOJ 2084: [Poi2010]Antisymmetry

Sol

Manacher.

\(O(n)\) Manacher很简单啊.改一改转移就可以了.

然后我WA了.一开始天真的认为id只会是奇数,然后就GG.

一组 Hack 数据

3
1 0 0

然后就跳过偶数的拓展...就过了...

Code

/**************************************************************
    Problem: 2084
    User: BeiYu
    Language: C++
    Result: Accepted
    Time:24 ms
    Memory:9100 kb
****************************************************************/
 
#include<cstdio>
#include<iostream>
  
using namespace std;
  
const int N = 500050;
typedef long long LL; 
  
int n,m;LL ans;
int ch[N<<1],p[N<<1];
  
inline int in(int x=0,char ch=getchar()){ while(ch>'9'||ch<'0') ch=getchar();
    while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();return x; }
inline char mychar(char ch=getchar()){ while(ch>'1'||ch<'0') ch=getchar();return ch; }
  
int main(){
//  freopen("in.in","r",stdin);
    n=in(),m=2*n+1;ch[0]=888,ch[1]=19,ch[m+1]=666;
    for(int i=1;i<=n;i++) ch[i<<1]=mychar()-'0',ch[i<<1|1]=19;
    int mx=0,id=0;
    for(int i=1;i<=m;i+=2){
        if(mx>i) p[i]=min(p[2*id-i],mx-i);
        else p[i]=1;
        while(ch[i+p[i]]+ch[i-p[i]]==1||ch[i+p[i]]+ch[i-p[i]]==38) p[i]++;
        if(i+p[i]>mx) mx=i+p[i],id=i;
    }
      
//  for(int i=1;i<=m;i++) printf("%2d ",ch[i]);cout<<endl;
//  for(int i=1;i<=m;i++) printf("%2d ",p[i]);cout<<endl;
    for(int i=1;i<=m;i+=2) ans+=(p[i]-1)>>1;
    printf("%lld\n",ans);
    return 0;
}