2015 Multi-University Training Contest 4 hdu 5338 ZZX and Permutations

ZZX and Permutations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 771 Accepted Submission(s): 243

Problem Description

ZZX likes permutations.

ZZX knows that a permutation can be decomposed into disjoint cycles(see https://en.wikipedia.org/wiki/Permutation#Cycle_notation). For example:
145632=(1)(35)(462)=(462)(1)(35)=(35)(1)(462)=(246)(1)(53)=(624)(1)(53)……
Note that there are many ways to rewrite it, but they are all equivalent.
A cycle with only one element is also written in the decomposition, like (1) in the example above.

Now, we remove all the parentheses in the decomposition. So the decomposition of 145632 can be 135462,462135,351462,246153,624153……

Now you are given the decomposition of a permutation after removing all the parentheses (itself is also a permutation). You should recover the original permutation. There are many ways to recover, so you should find the one with largest lexicographic order.

Input

First line contains an integer

Output

Output

Sample Input

1 4 5 6 3 2

1 2

Sample Output

4 6 2 5 1 3

2 1

Author

XJZX

Source

2015 Multi-University Training Contest 4

解题：线段树+set

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 300010;
 4 struct node {
 5     int lt,rt,lazy,maxv;
 6 } tree[maxn<<2];
 7 int d[maxn],val2index[maxn],n;
 8 void pushup(int v) {
 9     tree[v].maxv = max(tree[v<<1].maxv,tree[v<<1|1].maxv);
10 }
11 void pushdown(int v) {
12     if(tree[v].lazy > -1) {
13         tree[v<<1].lazy = tree[v<<1|1].lazy = tree[v].lazy;
14         tree[v<<1].maxv = tree[v<<1|1].maxv = tree[v].lazy;
15         tree[v].lazy = -1;
16     }
17 }
18 void build(int lt,int rt,int v) {
19     tree[v].lt = lt;
20     tree[v].rt = rt;
21     tree[v].lazy = -1;
22     if(lt == rt) {
23         tree[v].maxv = d[lt];
24         return;
25     }
26     int mid = (lt + rt)>>1;
27     build(lt,mid,v<<1);
28     build(mid+1,rt,v<<1|1);
29     pushup(v);
30 }
31 void update(int lt,int rt,int v) {
32     if(lt <= tree[v].lt && rt >= tree[v].rt) {
33         tree[v].lazy = tree[v].maxv = 0;
34         return;
35     }
36     pushdown(v);
37     if(lt <= tree[v<<1].rt) update(lt,rt,v<<1);
38     if(rt >= tree[v<<1|1].lt) update(lt,rt,v<<1|1);
39     pushup(v);
40 }
41 int query(int lt,int rt,int v) {
42     if(lt <= tree[v].lt && rt >= tree[v].rt) return tree[v].maxv;
43     pushdown(v);
44     int ret = 0;
45     if(lt <= tree[v<<1].rt) ret = max(ret,query(lt,rt,v<<1));
46     if(rt >= tree[v<<1|1].lt) ret = max(ret,query(lt,rt,v<<1|1));
47     pushup(v);
48     return ret;
49 }
50 set<int>st;
51 bool used[maxn];
52 int ret[maxn];
53 int main() {
54     int kase;
55     scanf("%d",&kase);
56     while(kase--) {
57         st.clear();
58         memset(used,false,sizeof used);
59         memset(ret,0,sizeof ret);
60         memset(d,0,sizeof d);
61         scanf("%d",&n);
62         for(int i = 1; i <= n; ++i) {
63             scanf("%d",d+i);
64             val2index[d[i]] = i;
65         }
66         build(0,n,1);
67         st.insert(0);
68         for(int i = 1; i <= n; ++i) {
69             if(ret[i]) continue;
70             int index = val2index[i],mx = 0;
71             if(!used[d[index+1]]) mx = max(d[index+1],mx);
72             auto it = st.lower_bound(index);
73             if(it != st.begin()) --it;
74             int val = query((*it),index,1);
75             mx = max(mx,val);
76             if(mx == d[index+1]) {
77                 used[d[index+1]] = true;
78                 ret[i] = d[index+1];
79                 update(index+1,index+1,1);
80                 continue;
81             }
82             ret[i] = mx;
83             for(int i = val2index[mx]; i < index; ++i) {
84                 ret[d[i]] = d[i+1];
85                 used[d[i]] = true;
86             }
87             update(val2index[mx],index,1);
88             used[d[index]] = true;
89             for(int i = val2index[mx]; i <= index; ++i)  st.insert(i);
90         }
91         for(int i = 1; i <= n; ++i)
92             printf("%d%c",ret[i],i==n?'
':' ');
93     }
94     return 0;
95 }

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