poj 3693 Maximum repetition substring

【分析】

参见论文

//poj 3693 Maximum repetition substring#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define ll long long#define M(a) memset(a,0,sizeof a)#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;const int mxn=200005;int n,m,T,len,ans,lef,rig;char s[mxn];int st[mxn][30],rank[mxn],height[mxn];int a[mxn],b[mxn],x[mxn],y[mxn],sa[mxn];inline bool comp(int i,int j,int l){    return y[i]==y[j]&&(i+l>len?-1:y[i+l])==(j+l>len?-1:y[j+l]);}inline void work(){    int i,j,k,p;m=200;    fo(i,0,m) b[i]=0;    fo(i,1,len) b[x[i]=a[i]]++;    fo(i,1,m) b[i]+=b[i-1];    for(i=len;i>=1;i--) sa[b[x[i]]--]=i;    for(k=1;k<=len;k<<=1)    {        p=0;        fo(i,len-k+1,len) y[++p]=i;        fo(i,1,len) if(sa[i]>k) y[++p]=sa[i]-k;        fo(i,0,m) b[i]=0;        fo(i,1,len) b[x[y[i]]]++;        fo(i,1,m) b[i]+=b[i-1];        for(i=len;i>=1;i--) sa[b[x[y[i]]]--]=y[i];        swap(x,y),p=2,x[sa[1]]=1;        fo(i,2,len)          x[sa[i]]=comp(sa[i-1],sa[i],k)?p-1:p++;        if(p>len) break;        m=p;    }    p=k=0;    fo(i,1,len) rank[sa[i]]=i;    for(i=1;i<=len;height[rank[i++]]=k)      for(k?k--:0,j=sa[rank[i]-1];a[i+k]==a[j+k];k++);}inline void init(){int i,j;fo(i,1,len) st[i][0]=height[i];fo(j,1,28)  fo(i,1,len) if((i+(1<<j)-1)<=len)    st[i][j]=min(st[i][j-1],st[i+(1<<j-1)][j-1]);}inline int query(int x,int y){if(x>y) swap(x,y);x++;int k=0;while(x+(1<<k+1)<=y) k++;return min(st[x][k],st[y-(1<<k)+1][k]);}int main(){int i,j,l,r,L;while(1){M(a);ans=lef=rig=0;scanf("%s",s+1);len=strlen(s+1);fo(i,1,len) a[i]=s[i];if(len==1 && s[1]=='#') break;printf("Case %d: ",++T);work(),init();fo(L,1,len/2){for(i=1;i*L+1<=len;i++){int x=(i-1)*L+1,y=i*L+1;if(s[x]!=s[y]) continue;r=y+query(rank[x],rank[y])-1;fo(j,0,L-1){if(x-j<1 || s[x-j]!=s[y-j]) break;int tmp=(r-(x-j)+1)/L;if(tmp>ans || (tmp==ans && rank[x-j]<rank[lef]))  lef=x-j,rig=lef+tmp*L-1,ans=tmp;}}}if(ans==0) printf("%c",s[sa[1]]);else fo(i,lef,rig) printf("%c",s[i]);printf("\n");}return 0;}/*xbcabcab*/