bzoj4829 [TJOI2017]dna（后缀数组）

luogu3763有题面可以去看的说。。大牛分站开o2优化了（没错我就是这样才过的。）后缀数组处理一下lcp，然后枚举三次就好了。bzoj上我的总是T。嗯，一定是姿势不对。。不过我看神犇们都是用SAM写的，贼快。我也要去学习一下。。

#include <cstdio>#include <cstring>#define N 200005char s[N];int tst,n,len1,len2,m,st[N][19],log[N];int rank[N<<1],rank1[N],sa[N],h[N],tmp[N],count[N];inline int swap(int &x,int &y){x^=y;y^=x;x^=y;}inline int min(int x,int y){return x<y?x:y;}inline int lcp(int x,int y){int l=rank[x],r=rank[y];if(l>r) swap(l,r);++l;int k=log[r-l+1];return min(st[l][k],st[r-(1<<k)+1][k]);}int main(){//freopen("dna.in","r",stdin);scanf("%d",&tst);log[0]=-1;for(int i=1;i<=N-5;++i) log[i]=log[i>>1]+1;while(tst--){int ans=0;scanf("%s",s+1);len1=strlen(s+1);m='Z'+1;s[len1+1]='Z';scanf("%s",s+len1+2);n=strlen(s+1);len2=strlen(s+len1+2);for(int i=1;i<=m;++i) count[i]=0;for(int i=1;i<=n;++i) count[s[i]]=1;for(int i=1;i<=m;++i) count[i]+=count[i-1];for(int i=1;i<=n;++i) rank[i]=count[s[i]];int k=0;for(int p=1;k!=n;p<<=1,m=k){for(int i=1;i<=m;++i) count[i]=0;for(int i=1;i<=n;++i) count[rank[i+p]]++;for(int i=1;i<=m;++i) count[i]+=count[i-1];for(int i=n;i>=1;--i) tmp[count[rank[i+p]]--]=i;for(int i=1;i<=m;++i) count[i]=0;for(int i=1;i<=n;++i) count[rank[tmp[i]]]++;for(int i=1;i<=m;++i) count[i]+=count[i-1];for(int i=n;i>=1;--i) sa[count[rank[tmp[i]]]--]=tmp[i];memcpy(rank1,rank,sizeof(rank1));rank[sa[1]]=k=1;for(int i=2;i<=n;++i){if(rank1[sa[i]]!=rank1[sa[i-1]]||rank1[sa[i]+p]!=rank1[sa[i-1]+p]) ++k;rank[sa[i]]=k;}}k=0;for(int i=1;i<=n;++i){if(rank[i]==1){h[1]=0;continue;}if(i==1||h[rank[i-1]]<=1) k=0;if(k) --k;while(s[i+k]==s[sa[rank[i]-1]+k]) ++k;h[rank[i]]=k;}for(int i=1;i<=n;++i) st[i][0]=h[i];for(int i=1;i<=log[n];++i)for(int j=1;j<=n;++j)if(j+(1<<i-1)<=n) st[j][i]=min(st[j][i-1],st[j+(1<<i-1)][i-1]);for(int i=1;i<=len1-len2+1;++i){int x=lcp(i,len1+2);if(x>=len2-3){ans++;continue;}++x;x+=lcp(i+x,len1+2+x);if(x>=len2-2){ans++;continue;}++x;x+=lcp(i+x,len1+2+x);if(x>=len2-1){ans++;continue;}++x;x+=lcp(i+x,len1+2+x);if(x>=len2){ans++;continue;}}printf("%d\n",ans);}return 0;}