bzoj 4278: [ONTAK2015]Tasowanie&bzoj 1692: [Usaco2007 Dec]队列变换 后缀数组+贪心

题意

给定两个数字串A和B，通过将A和B进行二路归并得到一个新的数字串T，请找到字典序最小的T。1<=n,m<=200000,1<=A[i],B[i]<=1000

分析

先把两个串放一起求sa，然后用两个指针维护两个串分别归并到哪一位，每次选字典序较大的即可。

证明看这里

代码

bzoj 4278

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=400005;

int n,m,s1
,s2
,rank[N*2],sa
,b
,d
,c
,s
;

void get_sa(int n,int m)
{
for (int i=1;i<=n;i++) b[s[i]]++;
for (int i=1;i<=m;i++) b[i]+=b[i-1];
for (int i=n;i>=1;i--) c[b[s[i]]--]=i;
int t=0;
for (int i=1;i<=n;i++)
{
if (s[c[i]]!=s[c[i-1]]) t++;
rank[c[i]]=t;
}
int j=1;
while (j<=n)
{
for (int i=1;i<=n;i++) b[i]=0;
for (int i=1;i<=n;i++) b[rank[i+j]]++;
for (int i=1;i<=n;i++) b[i]+=b[i-1];
for (int i=n;i>=1;i--) c[b[rank[i+j]]--]=i;
for (int i=1;i<=n;i++) b[i]=0;
for (int i=1;i<=n;i++) b[rank[i]]++;
for (int i=1;i<=n;i++) b[i]+=b[i-1];
for (int i=n;i>=1;i--) d[b[rank[c[i]]]--]=c[i];
t=0;
for (int i=1;i<=n;i++)
{
if (rank[d[i]]!=rank[d[i-1]]||rank[d[i]]==rank[d[i-1]]&&rank[d[i]+j]!=rank[d[i-1]+j]) t++;
c[d[i]]=t;
}
for (int i=1;i<=n;i++) rank[i]=c[i];
if (t==n) break;
j*=2;
}
for (int i=1;i<=n;i++) sa[rank[i]]=i;
}

int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++) scanf("%d",&s1[i]);
scanf("%d",&m);
for (int i=1;i<=m;i++) scanf("%d",&s2[i]);
for (int i=1;i<=n;i++) s[i]=s1[i];
s[n+1]=10000;
for (int i=1;i<=m;i++) s[i+n+1]=s2[i];
get_sa(n+m+1,10000);
int p=1,q=1;
while (p<=n||q<=m)
if (p<=n&&rank[p]<=rank[q+n+1]||q>m) printf("%d ",s1[p]),p++;
else printf("%d ",s2[q]),q++;
return 0;
}

bzoj 1692

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=60005;

int n,m,rank[N*2],sa
,b
,d
,c
;
char s
,s1
;

void get_sa(int n,int m)
{
for (int i=1;i<=n;i++) b[s[i]]++;
for (int i=1;i<=m;i++) b[i]+=b[i-1];
for (int i=n;i>=1;i--) c[b[s[i]]--]=i;
int t=0;
for (int i=1;i<=n;i++)
{
if (s[c[i]]!=s[c[i-1]]) t++;
rank[c[i]]=t;
}
int j=1;
while (j<=n)
{
for (int i=1;i<=n;i++) b[i]=0;
for (int i=1;i<=n;i++) b[rank[i+j]]++;
for (int i=1;i<=n;i++) b[i]+=b[i-1];
for (int i=n;i>=1;i--) c[b[rank[i+j]]--]=i;
for (int i=1;i<=n;i++) b[i]=0;
for (int i=1;i<=n;i++) b[rank[i]]++;
for (int i=1;i<=n;i++) b[i]+=b[i-1];
for (int i=n;i>=1;i--) d[b[rank[c[i]]]--]=c[i];
t=0;
for (int i=1;i<=n;i++)
{
if (rank[d[i]]!=rank[d[i-1]]||rank[d[i]]==rank[d[i-1]]&&rank[d[i]+j]!=rank[d[i-1]+j]) t++;
c[d[i]]=t;
}
for (int i=1;i<=n;i++) rank[i]=c[i];
if (t==n) break;
j*=2;
}
for (int i=1;i<=n;i++) sa[rank[i]]=i;
}

int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++) scanf("%s",s1+i);
for (int i=1;i<=n;i++) s[i]=s1[i];
s[n+1]='&';
for (int i=1;i<=n;i++) s[i+n+1]=s1[n-i+1];
get_sa(n*2+1,10000);
int p=1,q=n,tot=0;
while (p<=q)
{
if (rank[p]<=rank[n+1+n-q+1]) putchar(s1[p]),p++;
else putchar(s1[q]),q--;
tot=(tot+1)%80;
if (!tot) putchar('\n');
}
return 0;
}