BZOJ4310 跳蚤(后缀数组+二分答案)

  注意到答案一定是原串的子串,于是考虑造出SA,二分答案是第几小的子串。第k小子串很容易在SA上求出。之后计算使他成为最大子串至少要在几个位置切割,对每个字典序比答案大的后缀,找到所有合法切割位置(求lcp即可),就转化成了选最少的点使每个区间都包含至少一个点的经典问题。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 100010
#define ll long long
int n,m,sa[N],sa2[N],rk[N<<1],tmp[N<<1],h[N],cnt[N],l[N];
char s[N];
ll tot=0;
void make()
{
    int m=0;
    for (int i=1;i<=n;i++) cnt[rk[i]=s[i]]++,m=max(m,(int)s[i]);
    for (int i=1;i<=m;i++) cnt[i]+=cnt[i-1];
    for (int i=n;i>=1;i--) sa[cnt[rk[i]]--]=i;
    for (int k=1;k<=n;k<<=1)
    {
        int p=0;
        for (int i=n-k+1;i<=n;i++) sa2[++p]=i;
        for (int i=1;i<=n;i++) if (sa[i]>k) sa2[++p]=sa[i]-k;
        memset(cnt,0,sizeof(cnt));
        for (int i=1;i<=n;i++) cnt[rk[i]]++;
        for (int i=1;i<=m;i++) cnt[i]+=cnt[i-1];
        for (int i=n;i>=1;i--) sa[cnt[rk[sa2[i]]]--]=sa2[i];
        memcpy(tmp,rk,sizeof(rk));
        p=rk[sa[1]]=1;
        for (int i=2;i<=n;i++)
        {
            if (tmp[sa[i]]!=tmp[sa[i-1]]||tmp[sa[i]+k]!=tmp[sa[i-1]+k]) p++;
            rk[sa[i]]=p;
        }
        if (p==n) break;
        m=p;
    }
    for (int i=1;i<=n;i++)
    {
        h[i]=max(h[i-1]-1,0);
        while (s[i+h[i]]==s[sa[rk[i]-1]+h[i]]) h[i]++;
    }
}
bool check(ll k)
{
    int x=0;
    for (int i=1;i<=n;i++)
    if (k<=n-sa[i]+1-h[sa[i]]) {x=i;break;}
    else k-=n-sa[i]+1-h[sa[i]];
    k+=h[sa[x]];memset(l,0,sizeof(l));
    int lcp=k;if (k<n-sa[x]+1) l[sa[x]+lcp-1]=max(l[sa[x]+lcp-1],sa[x]);
    for (int i=x+1;i<=n;i++)
    {
        lcp=min(lcp,h[sa[i]]);
        if (!lcp) return 0;
        l[sa[i]+lcp-1]=max(l[sa[i]+lcp-1],sa[i]);
    }
    int t=1;
    for (int i=1;i<=n;i++)
    if (l[i])
    {
        t++;if (t>m) return 0;
        int x=i;while (x<n&&l[x+1]<=i) x++;
        i=x;
    }
    return 1;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4310.in","r",stdin);
    freopen("bzoj4310.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    m=read();
    scanf("%s",s+1);n=strlen(s+1);
    make();
    tot=1ll*n*(n+1)>>1;
    for (int i=1;i<=n;i++) tot-=h[i];
    ll l=1,r=tot,ans;
    while (l<=r)
    {
        ll mid=l+r>>1;
        if (check(mid)) r=mid-1,ans=mid;
        else l=mid+1;
    }
    for (int i=1;i<=n;i++)
    if (ans<=n-sa[i]+1-h[sa[i]]) {for (int j=sa[i];ans+h[sa[i]]>0;j++,ans--)printf("%c",s[j]);break;}
    else ans-=n-sa[i]+1-h[sa[i]];
    return 0;
}

 

posted @ 2018-10-25 00:19  Gloid  阅读(298)  评论(0编辑  收藏  举报