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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#402249 | #6419. Baby's First Suffix Array Problem | Kazemaru | RE | 0ms | 30172kb | C++14 | 3.0kb | 2024-04-30 09:56:47 | 2024-04-30 09:56:48 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define f(i,j,k) for(register int i=j;i<=k;++i)
#define g(i,j,k) for(register int i=j;i>=k;--i)
int n,m,s,l;
const int N=2e5;
struct SA{
const static int k=23;
int x[N],y[N],sa[N],rk[N*2],h[N];
int f[N][25],o[25],p[N];
inline void init(char*a,int n){
f(i,0,n*2)rk[i]=0;
f(i,1,n)rk[i]=a[i],sa[i]=i;
for(int k=1;k<n;k*=2){
f(i,1,n)x[i]=rk[i],y[i]=rk[i+k];
auto cmp=[&](int a,int b){return x[a]!=x[b]?x[a]<x[b]:y[a]<y[b];};
sort(sa+1,sa+n+1,cmp);
int m=rk[sa[1]]=1;
f(i,2,n)rk[sa[i]]=m+=cmp(sa[i-1],sa[i]);
}
int l=0;
f(i,1,n){
while(a[i+l]==a[sa[rk[i]-1]+l])++l;
h[rk[i]]=l;if(l)--l;
}
f(i,1,n)f[i][0]=h[i];
f(i,0,k)o[i]=i?o[i-1]*2:1;
f(i,0,n)p[i]=i?p[i-1]+(o[p[i-1]+1]<=i):0;
f(t,1,k)f(i,1,n)if(i+o[t]-1<=n)f[i][t]=min(f[i][t-1],f[i+o[t-1]][t-1]);
}
inline int ask(int l,int r){int t=p[r-l+1];return min(f[l][t],f[r-o[t]+1][t]);}
inline int lcp(int x,int y){return x==y?N:ask(min(x,y)+1,max(x,y));}
inline void lr(int x,int k,int&pl,int&pr){
x=rk[x];
int le=1,ri=x,mid;
while(le<ri){
mid=(le+ri)/2;
if(lcp(mid,x)<k)le=mid+1;
else ri=mid;
}pl=le;
le=x;ri=n;
while(le<ri){
mid=(le+ri+1)/2;
if(lcp(x,mid)<k)ri=mid-1;
else le=mid;
}pr=ri;
}
}S;
struct Kazemaru{
int c[N],n;
inline void init(int x){f(i,0,x)c[i]=0;n=x;}
inline void add(int x,int y){for(;x<=n;x+=-x&x)c[x]+=y;}
inline int sum(int x,int y=0){for(;x;x-=-x&x)y+=c[x];return y;}
}T;
char a[N];int b[N];
struct op{int l,r,t,p;}L[N],R[N];
vector<op>q[N],o[N];
inline bool operator<(op a,op b){return a.r>b.r;}
void cdq(int l,int r){
if(l==r)return;
int mid=(l+r)/2;
cdq(l,mid);cdq(mid+1,r);
//k<i<=r rk[i]>rk[k],lcp>=r-i+1
//max(k,r-lcp(k,mid))<i<=r -> L[k].l<R[i].p<=L[k].r;
//r<i+lcp(mid,i) ->L[k].r<R[i].r
int cl=0,cr=0,p=0;
f(i,l,mid){
int k=S.sa[i];
for(op e:o[i])L[++cl]=(op){max(k,e.r-S.lcp(i,mid)),e.r,0,e.p};
}
f(i,mid+1,r)R[++cr]=(op){0,S.sa[i]+S.lcp(mid,i),0,S.sa[i]};
sort(L+1,L+cl+1);sort(R+1,R+cr+1);
f(i,1,cl){
while(p<cr&&L[i].r<R[p+1].r)T.add(R[++p].p,1);
b[L[i].p]+=T.sum(L[i].r)-T.sum(L[i].l);
}
f(i,1,p)T.add(R[i].p,-1);
}
inline void doing(){
cin>>n>>m;
scanf("%s",a+1);
int r,k,p,pl,pr;
S.init(a,n);T.init(n);
f(i,1,n)q[i].clear(),o[i].clear();
// f(i,1,n)cout<<S.rk[i]<<" ";cout<<endl;
f(i,1,m){
cin>>l>>r>>k;b[i]=1;p=S.rk[k]-1;
//l<=i<k rk[i]<rk[k],lcp<r-k+1
S.lr(k,r-k+1,pl,pr);pr=min(pr,p);
q[l-1].push_back({1,p,-1,i});
q[k-1].push_back({1,p,1,i});
q[l-1].push_back({pl,pr,1,i});
q[k-1].push_back({pl,pr,-1,i});
//k<i<=r rk[i]<rk[k]
q[k].push_back({1,p-1,-1,i});
q[r].push_back({1,p-1,1,i});
//k<i<=r rk[i]>rk[k],lcp>=r-i+1
if(k<r)o[p+1].push_back({0,r,0,i});
}
f(i,1,n){
T.add(S.rk[i],1);
for(op e:q[i])b[e.p]+=e.t*(T.sum(e.r)-T.sum(e.l-1));
}
T.init(n);cdq(1,n);
f(i,1,m)printf("%lld\n",b[i]);
}
signed main(){
int t;
cin>>t;
while(t--)doing();
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 30172kb
input:
2 10 4 baaabbabba 2 8 3 1 1 1 2 3 2 2 5 4 20 3 cccbccbadaacbbbcccab 14 17 16 3 20 17 17 20 18
output:
2 1 2 3 4 15 3
result:
ok 7 numbers
Test #2:
score: -100
Runtime Error
input:
8994 10 10 abaabbaabb 2 8 3 1 8 5 6 9 6 9 9 9 5 10 7 5 7 7 7 8 7 5 6 6 1 9 9 3 9 5 2 1 ab 1 2 1 8 6 bbabaabb 3 7 7 5 7 7 1 5 1 1 4 3 3 5 3 5 5 5 10 3 aababbaaab 3 4 4 6 9 8 4 6 6 7 3 babbaab 5 6 5 3 6 6 1 6 6 9 3 baaaabbba 2 5 2 8 9 8 1 4 4 9 2 babaababa 2 4 4 2 6 3 2 3 ba 1 2 2 1 2 1 2 2 2 10 2 aba...