[后缀数组 贪心] BZOJ 4278 [ONTAK2015]Tasowanie

两个指针 显然小的那个先放 如果一样 比后一个 再一样 再后 然后就转化成比较后缀的字典序了

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=400005;int n1,n2;int n,a[N];int t1[N],t2[N],c[N],sa[N];int rank[N];inline void SA(int *a,int m){  int *x=t1,*y=t2;  for (int i=1;i<=m;i++) c[i]=0;  for (int i=1;i<=n;i++) c[x[i]=a[i]]++;  for (int i=1;i<=m;i++) c[i]+=c[i-1];  for (int i=n;i;i--) sa[c[x[i]]--]=i;  for (int k=1;k<=n;k<<=1){    int p=0;    for (int i=n-k+1;i<=n;i++) y[++p]=i;    for (int i=1;i<=n;i++) if (sa[i]>k) y[++p]=sa[i]-k;    for (int i=1;i<=m;i++) c[i]=0;    for (int i=1;i<=n;i++) c[x[y[i]]]++;    for (int i=1;i<=m;i++) c[i]+=c[i-1];    for (int i=n;i;i--) sa[c[x[y[i]]]--]=y[i];    swap(x,y);    x[sa[1]]=1; p=1;    for (int i=2;i<=n;i++)      x[sa[i]]=(y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+k])?p:++p;    if (p>=n) break;    m=p;  }}int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(n1); for (int i=1;i<=n1;i++) read(a[i]);  a[n1+1]=1001;  read(n2); for (int i=1;i<=n2;i++) read(a[i+n1+1]);  n=n1+1+n2;  SA(a,1001);  for (int i=1;i<=n;i++) rank[sa[i]]=i;  int p=1,q=1;  while (p<=n1 || q<=n2)    if (p>n1)      printf("%d ",a[(q++)+n1+1]);    else if (q>n2)      printf("%d ",a[p++]);    else if (rank[p]<rank[q+n1+1])      printf("%d ",a[p++]);    else      printf("%d ",a[(q++)+n1+1]);  return 0;}