每次从等待序列的头或尾拿出一个放到答案序列的末尾,那么每次贪心比较头和尾的字典序大小即可……

  TAT贪心很好想,但是我一开始没想到是可以直接比较字符串大小……而是一位一位判的,WA了……

  膜拜了zyf的做法TAT

 1 /**************************************************************
 2     Problem: 1692
 3     User: Tunix
 4     Language: C++
 5     Result: Accepted
 6     Time:248 ms
 7     Memory:3616 kb
 8 ****************************************************************/
 9  
10 //BZOJ 1640
11 #include<vector>
12 #include<cstdio>
13 #include<cstring>
14 #include<cstdlib>
15 #include<iostream>
16 #include<algorithm>
17 #define rep(i,n) for(int i=0;i<n;++i)
18 #define F(i,j,n) for(int i=j;i<=n;++i)
19 #define D(i,j,n) for(int i=j;i>=n;--i)
20 #define pb push_back
21 using namespace std;
22 inline int getint(){
23     int v=0,sign=1; char ch=getchar();
24     while(ch<'0'||ch>'9'){ if (ch=='-') sign=-1; ch=getchar();}
25     while(ch>='0'&&ch<='9'){ v=v*10+ch-'0'; ch=getchar();}
26     return v*sign;
27 }
28 const int N=1e5+10,INF=~0u>>2;
29 typedef long long LL;
30 /******************tamplate*********************/
31 int n,m,a[N];
32 inline bool cmp(int *r,int a,int b,int l){
33     return r[a]==r[b] && r[a+l]==r[b+l];
34 }
35 int wa[N],wb[N],c[N],sa[N],rank[N];
36 void DA(int *s,int *sa,int n,int m){
37     int i,j,p,*x=wa,*y=wb;
38     rep(i,m) c[i]=0;
39     rep(i,n) c[x[i]=s[i]]++;
40     F(i,1,m-1) c[i]+=c[i-1];
41     D(i,n-1,0) sa[--c[x[i]]]=i;
42     for(j=1,p=1;p<n;j<<=1,m=p){
43         for(p=0,i=n-j;i<n;i++) y[p++]=i;
44         rep(i,n) if (sa[i]>=j) y[p++]=sa[i]-j;
45  
46         rep(i,m) c[i]=0;
47         rep(i,n) c[x[y[i]]]++;
48         F(i,1,m-1) c[i]+=c[i-1];
49         D(i,n-1,0) sa[--c[x[y[i]]]]=y[i];
50         swap(x,y); x[sa[0]]=0; p=1;
51         F(i,1,n-1)
52             x[sa[i]]=cmp(y,sa[i-1],sa[i],j) ? p-1 : p++;
53     }
54 }
55 int main(){
56 #ifndef ONLINE_JUDGE
57     freopen("1640.in","r",stdin);
58     freopen("1640.out","w",stdout);
59 #endif
60     n=getint(); m=0; int tmp=0;
61     char s[5];
62     rep(i,n){
63         scanf("%s",s);
64         a[i]=s[0]-'A'+1;
65     }
66     int tot=n*2+1;
67     F(i,1,n) a[n+i]=a[n-i];
68     DA(a,sa,tot+1,28);
69     F(i,1,tot) rank[sa[i]]=i;
70     int l=0,r=n+1;
71     while(l+r-n-1<n){
72         if (rank[l]<rank[r]) putchar(a[l++]+'A'-1);
73         else putchar(a[r++]+'A'-1);
74         if (!((l+r-n-1)%80)) puts("");
75     }
76     return 0;
77 }
View Code(1692)

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