[BZOJ3230]相似子串(后缀数组)
显然可以通过后缀数组快速找到询问的两个串分别是什么，然后正反各建一个后缀数组来求两个串的LCP和LCS即可。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rep(i,l,r) for (int i=(l); i<=(r); i++) 5 typedef long long ll; 6 using namespace std; 7 8 const int N=100010,inf=1e9; 9 char s[N]; 10 ll ans,x,y; 11 int n,Q,lg[N]; 12 struct P{ int x,y; }; 13 14 struct SA{ 15 ll sm[N]; 16 char s[N]; 17 int x[N],y[N],c[N],sa[N],rk[N],h[N],mn[N][20]; 18 19 bool Cmp(int a,int b,int l){ return a+l<=n && b+l<=n && y[a]==y[b] && y[a+l]==y[b+l]; } 20 21 void build(int m){ 22 rep(i,0,m) c[i]=0; 23 rep(i,1,n) c[x[i]=s[i]-'a'+1]++; 24 rep(i,1,m) c[i]+=c[i-1]; 25 for (int i=n; i; i--) sa[c[x[i]]--]=i; 26 for (int k=1,p=0; p<n; k<<=1,m=p){ 27 p=0; 28 rep(i,n-k+1,n) y[++p]=i; 29 rep(i,1,n) if (sa[i]>k) y[++p]=sa[i]-k; 30 rep(i,0,m) c[i]=0; 31 rep(i,1,n) c[x[y[i]]]++; 32 rep(i,1,m) c[i]+=c[i-1]; 33 for (int i=n; i; i--) sa[c[x[y[i]]]--]=y[i]; 34 rep(i,1,n) y[i]=x[i]; x[sa[p=1]]=1; 35 rep(i,2,n) x[sa[i]]=Cmp(sa[i],sa[i-1],k) ? p : ++p; 36 } 37 } 38 39 void init(){ 40 rep(i,1,n) rk[sa[i]]=i; 41 int k=0; 42 rep(i,1,n){ 43 for (int j=sa[rk[i]-1]; i+k<=n && j+k<=n && s[i+k]==s[j+k]; k++); 44 h[rk[i]]=k; if (k) k--; 45 } 46 rep(i,1,n) sm[i]=sm[i-1]+n-sa[i]+1-h[i],mn[i][0]=h[i]; 47 rep(j,1,18) rep(i,1,n-(1<<j)+1) mn[i][j]=min(mn[i][j-1],mn[i+(1<<(j-1))][j-1]); 48 } 49 50 int que(int l,int r){ 51 if (l==r) return inf; 52 l++; int t=lg[r-l+1]; return min(mn[l][t],mn[r-(1<<t)+1][t]); 53 } 54 55 P get(ll k){ 56 int L=1,R=n; int ans1,ans2; 57 if (sm[n]<k) return (P){-1,-1}; 58 while (L<=R){ 59 int mid=(L+R)>>1; 60 if (sm[mid]>=k) ans1=mid,ans2=n-sa[mid]+1-(sm[mid]-k),R=mid-1; else L=mid+1; 61 } 62 return (P){ans1,ans2}; 63 } 64 }S1,S2; 65 66 int main(){ 67 freopen("bzoj3230.in","r",stdin); 68 freopen("bzoj3230.out","w",stdout); 69 scanf("%d%d%s",&n,&Q,s+1); 70 rep(i,2,n) lg[i]=lg[i>>1]+1; 71 rep(i,1,n) S1.s[i]=S2.s[n-i+1]=s[i]; 72 S1.build(30); S2.build(30); S1.init(); S2.init(); 73 rep(i,1,Q){ 74 scanf("%lld%lld",&x,&y); ans=0; 75 if (S1.sm[n]<y){ puts("-1"); continue; } 76 P xx=S1.get(x),yy=S1.get(y); 77 ll k=min(S1.que(xx.x,yy.x),min(xx.y,yy.y)); ans+=k*k; 78 xx.x=S2.rk[n-S1.sa[xx.x]+2-xx.y]; yy.x=S2.rk[n-S1.sa[yy.x]+2-yy.y]; 79 k=min(S2.que(min(xx.x,yy.x),max(xx.x,yy.x)),min(xx.y,yy.y)); ans+=k*k; 80 printf("%lld\n",ans); 81 } 82 return 0; 83 }
