【poj3581】Sequence 后缀数组
2016-02-26 18:41
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Description
Given a sequence, {A1, A2, …, An} which is guaranteed A1 > A2, …, An, you are to cut it into three sub-sequences and reverse them separately to form a new one which is the smallest possible sequence in alphabet order.The alphabet order is defined as follows: for two sequence {A1, A2, …, An} and {B1, B2, …, Bn}, we say {A1, A2, …, An} is smaller than {B1, B2, …, Bn} if and only if there exists such i ( 1 ≤ i ≤ n) so that we have Ai < Bi and Aj = Bj for each j < i.
Input
The first line contains n. (n ≤ 200000)The following n lines contain the sequence.
Output
output n lines which is the smallest possible sequence obtained.Sample Input
5 10 1 2 3 4
Sample Output
1 10 2 4 3
Hint
{10, 1, 2, 3, 4} -> {10, 1 | 2 | 3, 4} -> {1, 10, 2, 4, 3}Source
POJ Founder Monthly Contest – 2008.04.13, Yao Jinyu把串分成三份,把每份翻转,使得最终串的字典序最小。三份不能有空串。
这个后缀数组好蛋疼…代码恶心,最后main函数我抄的书233
因为第一个元素最大,所以第一段很好找,找反串的最小字典序的后缀即可。
然后剩下就不能这么简单地分了,可以考虑:
若把后面分为两段,然后把它们翻转,使得字典序最小,相当于把串复制一份接后面然后求反串的最小字典序后缀。
例如串是这样:ABCD
扩大一倍成这样:ABCDABCD
反串:DCBADCBA
若解是把AB和CD翻转,那么BADC的字典序最小,对应的在反串里的后缀就是BADCBA,取前缀即可。
要注意合法,要注意细节
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int SZ = 1000010; int n; int sa[SZ],rank[SZ],k = 1,tmp[SZ]; bool cmp_sa(int i,int j) { if(rank[i] != rank[j]) return rank[i] < rank[j]; else { int x = i + k <= n ? rank[i + k] : -1; int y = j + k <= n ? rank[j + k] : -1; return x < y; } } void get_sa(int s[],int n) { for(int i = 0;i <= n;i ++) { sa[i] = i; rank[i] = i == n ? -1 : s[i]; } for(k = 1;k <= n;k <<= 1) { sort(sa,sa + 1 + n,cmp_sa); tmp[sa[0]] = 0; for(int i = 1;i <= n;i ++) tmp[sa[i]] = tmp[sa[i - 1]] + (cmp_sa(sa[i - 1],sa[i]) ? 1 : 0); for(int i = 0;i <= n;i ++) rank[i] = tmp[i]; } } int num[SZ],s[SZ]; int tmp2[SZ]; void copy(int num[],int l,int r,int s[]) { for(int i = l;i < r;i ++) s[i - l] = num[i]; } void rev(int s[],int l,int r) { for(int i = l;i < r;i ++) { tmp2[i] = s[r - i - 1 + l]; // cout<<tmp2[i]<<" "<<i<<endl; } // puts(""); for(int i = l;i < r;i ++) s[i] = tmp2[i]; // for(int i = 0;i < n;i ++) printf("%d ",num[i]); puts(""); } int main() { scanf("%d",&n); for(int i = 0;i < n;i ++) scanf("%d",&num[i]); copy(num,0,n,s); rev(s,0,n); get_sa(s,n); int p1; for(int i = 0;i < n;i ++) { p1 = n - sa[i]; if(p1 >= 1 && n - p1 >= 2) break; } int m = n - p1; copy(num,p1,n,s); copy(num,p1,n,s + m); rev(s,0,m * 2); get_sa(s,m * 2); int p2; for(int i = 0;i <= m * 2;i ++) { p2 = p1 + m - sa[i]; if(p2 - p1 >= 1 && n - p2 >= 1) break; } // for(int i = 0;i < n;i ++) printf("%d ",num[i]); // cout<<endl<<p1<<" 233333 "<<p2<<endl; rev(num,0,p1); rev(num,p1,p2); rev(num,p2,n); for(int i = 0;i < n;i ++) printf("%d\n",num[i]); return 0; } /* 2 3 4 2 3 4 4 3 2 4 3 2 0 1 2 3 4 5 8 5 0 3 1 2 3 1 4 4 1 3 2 1 3 0 5 3 1 2 3 1 4 3 1 2 3 1 4 4 1 3 2 1 3 4 1 3 2 1 3 5 5 4 3 2 1 1 2 1 2 */
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