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.
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 t, the number of test cases.
Then t testcases follow. In each testcase:
First line contains an integer n, the size of the permutation.
Second line contains n space-separated integers, the decomposition after removing parentheses.
n≤105. There are 10 testcases satisfying n≤105, 200 testcases satisfying n≤1000.
Then t testcases follow. In each testcase:
First line contains an integer n, the size of the permutation.
Second line contains n space-separated integers, the decomposition after removing parentheses.
n≤105. There are 10 testcases satisfying n≤105, 200 testcases satisfying n≤1000.
Output
Output n space-separated numbers in a line for each testcase.
Don't output space after the last number of a line.
Don't output space after the last number of a line.
Sample Input
2
6
1 4 5 6 3 2
2
1 2
Sample Output
4 6 2 5 1 3
2 1
Author
XJZX
Source
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?'\n':' '); 93 } 94 return 0; 95 }
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