hdu5157 Harry and magic string 回文树

统计一个字符串中不相交的回文串的对数。

枚举以i为右端点的回文串,乘以在区间[i+1,n]的回文串个数,累加即可。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
const int maxn=200100;
const int INF=1e9+10;

char s[maxn],t[maxn];
int ls,lt;

struct PalinTree
{
    int ch[maxn][26],f[maxn];
    int n,tot,last;
    int len[maxn];
    int s[maxn];
    ll num[maxn];
    ll cnt[maxn];
    ll cnt2[maxn];
    int newnode(int l)
    {
        MS0(ch[tot]);
        len[tot]=l;
        return tot++;
    }
    void init()
    {
        tot=0;
        newnode(0);
        newnode(-1);
        n=last=0;
        s[n]=-1;f[0]=1;
        num[0]=0;num[1]=-1;
        cnt[0]=0;
        cnt2[0]=0;
    }
    int get_fail(int x)
    {
        while(s[n-len[x]-1]!=s[n]) x=f[x];
        return x;
    }
    void add(int c)
    {
        c-='a';
        s[++n]=c;
        last=get_fail(last);
        if(!ch[last][c]){
            int cur=newnode(len[last]+2);
            f[cur]=ch[get_fail(f[last])][c];
            num[cur]=num[f[cur]]+1;
            ch[last][c]=cur;
        }
        last=ch[last][c];
        cnt[n]=cnt[n-1]+num[last];
        cnt2[n]=num[last];
    }
    void build(char *str)
    {
        init();
        int ls=strlen(str);
        REP(i,0,ls-1) add(str[i]);
    }
};PalinTree ps,pt;

int main()
{
    freopen("in.txt","r",stdin);
    while(~scanf("%s",s)){
        ls=lt=strlen(s);
        REP(i,0,ls-1) t[i]=s[ls-i-1];
        ps.build(s);
        pt.build(t);
        ll ans=0;
        REP(i,1,ls-1){
            ans+=1LL*ps.cnt[i]*pt.cnt2[ls-i];
        }
        printf("%I64d\n",ans);
    }
    return 0;
}
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posted @ 2016-04-07 13:47  __560  阅读(269)  评论(0编辑  收藏  举报