题意:
给你一串数,让你找重复出现不少于k次的子串的最长长度,重复出现可重叠
题解:
老套路,还是二分答案,然后用height数组来check答案
1 #include<cstdio> 2 #include<algorithm> 3 #define F(i,a,b) for(int i=a;i<=b;i++) 4 using namespace std; 5 6 namespace suffixarray{ 7 #define FN(n) for(int i=0;i<n;i++) 8 const int N =2E5+7; 9 int rnk[N],sa[N],height[N],c[N],s[N]; 10 void getsa(int n,int m,int *x=rnk,int *y=height){ 11 FN(m)c[i]=0;FN(n)c[x[i]=s[i]]++;FN(m)c[i+1]+=c[i]; 12 for(int i=n-1;i>=0;i--)sa[--c[x[i]]]=i; 13 for(int k=1,p;p=0,k<=n;k=p>=n?N:k<<1,m=p){ 14 for(int i=n-k;i<n;i++)y[p++]=i; 15 FN(n)if(sa[i]>=k)y[p++]=sa[i]-k; 16 FN(m)c[i]=0;FN(n)c[x[y[i]]]++;FN(m)c[i+1]+=c[i]; 17 for(int i=n-1;i>=0;i--)sa[--c[x[y[i]]]]=y[i]; 18 swap(x,y),p=1,x[sa[0]]=0; 19 for(int i=1;i<n;i++) 20 x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++; 21 } 22 FN(n)rnk[sa[i]]=i; 23 for(int i=0,j,k=0;i<n-1;height[rnk[i++]]=k) 24 for(k=k?k-1:k,j=sa[rnk[i]-1];s[i+k]==s[j+k];k++); 25 } 26 } 27 using namespace suffixarray; 28 int n,k,a[N],ed; 29 30 inline int getid(int x){return lower_bound(a+1,a+1+ed,x)-a;} 31 32 bool check(int x,int cnt=1) 33 { 34 F(i,2,n)if(height[i]>=x){if(++cnt>=k)return 1;}else cnt=1; 35 return 0; 36 } 37 38 int main() 39 { 40 while(~scanf("%d%d",&n,&k)) 41 { 42 F(i,1,n)scanf("%d",a+i),s[i-1]=a[i]; 43 sort(a+1,a+1+n),ed=1; 44 F(i,2,n)if(a[i]!=a[ed])a[++ed]=a[i]; 45 F(i,0,n-1)s[i]=getid(s[i]); 46 s[n]=0,getsa(n+1,ed+1); 47 int l=1,r=n,mid,ans=0; 48 while(l<=r)mid=(l+r)>>1,check(mid)?l=(ans=mid)+1:r=mid-1; 49 printf("%d\n",ans); 50 } 51 return 0; 52 }
时间: 2024-10-21 05:42:57