hdu5338 ZZX and Permutations(贪心、线段树)

转载请注明出处： http://www.cnblogs.com/fraud/           ——by fraud

 

ZZX and Permutations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 181    Accepted Submission(s): 38


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 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.
 


Output
Output n space-separated numbers in a line for each testcase.
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

和题解的解法几乎一致，然而却在比赛结束后才意识到自己跪在了一个傻逼的地方，真是2333333

贪心，先放第一个位置，放尽可能大的值，然后依次往后，对于每个值，看其后面的一个位置，或者前面没有使用过的连续区间

  1 //#####################
  2 //Author:fraud
  3 //Blog: http://www.cnblogs.com/fraud/
  4 //#####################
  5 //#pragma comment(linker, "/STACK:102400000,102400000")
  6 #include <iostream>
  7 #include <sstream>
  8 #include <ios>
  9 #include <iomanip>
 10 #include <functional>
 11 #include <algorithm>
 12 #include <vector>
 13 #include <string>
 14 #include <list>
 15 #include <queue>
 16 #include <deque>
 17 #include <stack>
 18 #include <set>
 19 #include <map>
 20 #include <cstdio>
 21 #include <cstdlib>
 22 #include <cmath>
 23 #include <cstring>
 24 #include <climits>
 25 #include <cctype>
 26 using namespace std;
 27 #define XINF INT_MAX
 28 #define INF 0x3FFFFFFF
 29 #define MP(X,Y) make_pair(X,Y)
 30 #define PB(X) push_back(X)
 31 #define REP(X,N) for(int X=0;X<N;X++)
 32 #define REP2(X,L,R) for(int X=L;X<=R;X++)
 33 #define DEP(X,R,L) for(int X=R;X>=L;X--)
 34 #define CLR(A,X) memset(A,X,sizeof(A))
 35 #define IT iterator
 36 typedef long long ll;
 37 typedef pair<int,int> PII;
 38 typedef vector<PII> VII;
 39 typedef vector<int> VI;
 40 int Scan() {
 41     int res=0, ch;
 42     while(ch=getchar(), ch<'0'||ch>'9');
 43     res=ch-'0';
 44     while((ch=getchar())>='0'&&ch<='9')
 45         res=res*10+ch-'0';
 46     return res;
 47 }
 48 void Out(int a) {
 49     if(a>9)
 50         Out(a/10);
 51     putchar(a%10+'0');
 52 }
 53 int a[100010];
 54 int vis[100010];
 55 int pos[100010];
 56 int fa[100010];
 57 int dp[600010];
 58 int ans[100010];
 59 set<int> ms;
 60 int n;
 61 set<int>::IT it;
 62 int find(int x){
 63     if(fa[x]!=x)fa[x] = find(fa[x]);
 64     return fa[x];
 65 }
 66 void push_up(int cur){
 67     dp[cur] = max(dp[cur<<1],dp[cur<<1|1]);
 68 }
 69 void build(int l,int r,int cur){
 70     if(l==r){
 71         dp[cur] = a[l];
 72         return;
 73     }
 74     int mid = (l+r)>>1;
 75     build(l,mid,cur<<1);
 76     build(mid+1,r,cur<<1|1);
 77     push_up(cur);
 78 }
 79 void update(int l,int r,int cur,int x,int inc){
 80     if(l==r&&l==x){
 81         dp[cur] = inc;
 82         return;
 83     }
 84     int mid = (l+r)>>1;
 85     if(x<=mid)update(l,mid,cur<<1,x,inc);
 86     else update(mid+1,r,cur<<1|1,x,inc);
 87     push_up(cur);
 88 }
 89 void update(int x,int num){
 90     update(1,n,1,x,num);
 91 }
 92 int query(int l,int r,int lx,int rx,int cur){
 93     if(l>=lx&&r<=rx)return dp[cur];
 94     if(lx>r||rx<l)return 0;
 95     int mid = (l+r)>>1;
 96     return max(query(l,mid,lx,rx,cur<<1),query(mid+1,r,lx,rx,cur<<1|1));
 97 }
 98 int query(int l,int r){
 99     return query(1,n,l,r,1);
100 }    
101 int main(){
102     int t;
103     t = Scan();
104     while(t--){
105         n = Scan();
106         for(int i = 1;i<=n;i++){
107             a[i] = Scan();
108             vis[i] = 0;
109             pos[a[i]] = i;
110             fa[i] = i;
111         }
112         for(int i=0;i<=3*n;i++)dp[i] = 0;
113         build(1,n,1);
114         ms.clear();
115         ms.insert(0);
116         ms.insert(-INF);
117         for(int i=1;i<=n;i++){
118             if(vis[i])continue;
119             int l;
120             int posi = pos[i];
121             int maxx = find(i);
122             int p =pos[maxx];
123             if(posi+1<=n&&!vis[a[posi+1]]){
124                 if(a[posi+1]>maxx){
125                     p = posi+1;
126                     maxx = a[p];
127                 }
128             }
129             if(posi>1){
130                 it = ms.upper_bound(-posi);
131                 l = *it;
132                 l = -l;
133                 l++;
134                 int tmp = query(l,posi);
135                 //tmp = find(tmp);
136                 //cout<<l<<" "<<tmp<<endl;
137                 if(tmp>maxx){
138                     maxx = tmp;
139                     p = pos[tmp];
140                 }
141             }
142             if(p==posi){
143                 ans[i] = i;
144                 vis[i] = 1;
145                 ms.insert(-posi);
146             }else if(p==posi+1){
147                 fa[maxx] = fa[i];
148                 //ans[posi] = maxx;
149                 update(p,0);
150             }else{
151                 ans[i] = a[p];
152                 vis[a[p]] = 1;
153                 for(int j=p+1;j<=posi;j++){
154                     ans[a[j-1]] = a[j];
155                     vis[a[j]] = 1;
156                 }
157                 ms.insert(-posi);
158             }
159         /*    cout<<i<<" "<<p<<" "<<maxx<<endl;
160             for(int i=1;i<=n;i++){
161                 cout<<ans[i]<<" ";
162             }
163             cout<<endl;*/
164         }
165         for(int i=1;i<=n;i++){
166             if(i!=1)putchar(' ');
167             Out(ans[i]);
168         }
169         puts("");
170     }
171     return 0;
172 }
173                 
174

 

posted on 2015-07-30 20:51  xyiyy  阅读(290)  评论(0)  编辑  收藏  举报

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