E - Young Maids
Time limit : 2sec / Memory limit : 256MB
Score : 800 points
Problem Statement
Let N be a positive even number.
We have a permutation of (1,2,…,N), p=(p1,p2,…,pN). Snuke is constructing another permutation of (1,2,…,N), q, following the procedure below.
First, let q be an empty sequence. Then, perform the following operation until p becomes empty:
- Select two adjacent elements in p, and call them x and y in order. Remove x and y from p (reducing the length of p by 2), and insert x and y, preserving the original order, at the beginning of q.
When p becomes empty, q will be a permutation of (1,2,…,N).
Find the lexicographically smallest permutation that can be obtained as q.
Constraints
- N is an even number.
- 2≤N≤2×105
- p is a permutation of (1,2,…,N).
Input
Input is given from Standard Input in the following format:
N p1 p2 … pN
Output
Print the lexicographically smallest permutation, with spaces in between.
Sample Input 1
Copy
4 3 2 4 1
Sample Output 1
Copy
3 1 2 4
用下标从0开始的数组记录数，显然最终能作为开头的一定是下标为偶数的。由于是字典序，很容易想到贪心的想法，即从第一位开始，每一位取尽可能小的。暴力寻找最小的会使整个算法为n^2，复杂度过大，故采用线段树进行优化，维护区间奇数下标、偶数下标的数的最小值，使复杂度降到nlogn。
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <vector> 5 #include <set> 6 #include <map> 7 #include <string> 8 #include <cstring> 9 #include <stack> 10 #include <queue> 11 #include <cmath> 12 #include <ctime> 13 #include <utility> 14 using namespace std; 15 #define REP(I,N) for (I=0;I<N;I++) 16 #define rREP(I,N) for (I=N-1;I>=0;I--) 17 #define rep(I,S,N) for (I=S;I<N;I++) 18 #define rrep(I,S,N) for (I=N-1;I>=S;I--) 19 #define FOR(I,S,N) for (I=S;I<=N;I++) 20 #define rFOR(I,S,N) for (I=N;I>=S;I--) 21 typedef unsigned long long ull; 22 typedef long long ll; 23 //const int INF=0x3f3f3f3f; 24 const int INF=1e9; 25 const ll INFF=0x3f3f3f3f3f3f3f3fll; 26 const ll M=1e9+7; 27 const ll maxn=1e5+7; 28 const int MAXN=1005; 29 const int MAX=2e5+5; 30 const int MAX_N=MAX; 31 const ll MOD=1e9+7; 32 //const double eps=0.00000001; 33 //ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} 34 template<typename T>inline T abs(T a) {return a>0?a:-a;} 35 inline ll powMM(ll a,ll b){ 36 ll ret=1; 37 a%=M; 38 // b%=M; 39 while (b){ 40 if (b&1) ret=ret*a%M; 41 b>>=1; 42 a=a*a%M; 43 } 44 return ret; 45 } 46 int mina[MAX<<2][2]; 47 int a[MAX],lo[MAX],n; 48 void init(int l,int r,int k) 49 { 50 mina[k][0]=mina[k][1]=INF; 51 if(l==r) 52 return; 53 init(l,(l+r)/2,2*k); 54 init((l+r)/2+1,r,2*k+1); 55 } 56 void build(int l,int r,int k)//左闭右闭 57 { 58 if(l==r) 59 { 60 scanf("%d",&a[l]); 61 lo[a[l]]=l; 62 mina[k][l%2]=a[l];return; 63 } 64 int mid=(l+r)/2; 65 build(l,mid,2*k); 66 build(mid+1,r,2*k+1); 67 mina[k][0]=min(mina[2*k][0],mina[2*k+1][0]); 68 mina[k][1]=min(mina[2*k][1],mina[2*k+1][1]); 69 } 70 int query(int ql,int qr,int l,int r,int k,int st) 71 { 72 if(l>=ql&&r<=qr) 73 return mina[k][st]; 74 int mid=(l+r)/2; 75 if(mid<ql) 76 return query(ql,qr,mid+1,r,2*k+1,st); 77 else if(mid>=qr) 78 return query(ql,qr,l,mid,2*k,st); 79 else 80 return min(query(ql,qr,l,mid,2*k,st),query(ql,qr,mid+1,r,2*k+1,st)); 81 } 82 struct node 83 { 84 int l,r,xiao,status; 85 bool operator <(node b)const 86 { 87 return xiao>b.xiao; 88 } 89 node(int x=0,int y=0,int z=0,int s=0):l(x),r(y),xiao(z),status(s){} 90 }; 91 priority_queue<node>que; 92 int an[MAX],cnt; 93 int main() 94 { 95 cnt=0; 96 scanf("%d",&n); 97 init(0,n-1,1); 98 build(0,n-1,1); 99 node tem; 100 tem.l=0;tem.r=n-1;tem.xiao=query(0,n-1,0,n-1,1,0);tem.status=0; 101 que.push(tem); 102 while(!que.empty()) 103 { 104 tem=que.top();que.pop(); 105 int now=lo[tem.xiao]; 106 int nxt=query(now+1,tem.r,0,n-1,1,1^tem.status); 107 int to=lo[nxt]; 108 an[cnt++]=tem.xiao;an[cnt++]=nxt; 109 if(now>tem.l) 110 que.push(node(tem.l,now-1,query(tem.l,now-1,0,n-1,1,tem.status),tem.status)); 111 if(to-now>1)//中间的区间奇偶性会变化 112 que.push(node(now+1,to-1,query(now+1,to-1,0,n-1,1,tem.status^1),tem.status^1)); 113 if(tem.r>to) 114 que.push(node(to+1,tem.r,query(to+1,tem.r,0,n-1,1,tem.status),tem.status)); 115 } 116 for(int i=0;i<cnt;i++) 117 printf("%d ",an[i]); 118 return 0; 119 }