On February, 30th n students came in the Center for Training Olympiad Programmers (CTOP) of the Berland State University. They came one by one, one after another. Each of them went in, and before sitting down at his desk, greeted with those who were present in the room by shaking hands. Each of the students who came in stayed in CTOP until the end of the day and never left.
At any time any three students could join together and start participating in a team contest, which lasted until the end of the day. The team did not distract from the contest for a minute, so when another student came in and greeted those who were present, he did not shake hands with the members of the contest writing team. Each team consisted of exactly three students, and each student could not become a member of more than one team. Different teams could start writing contest at different times.
Given how many present people shook the hands of each student, get a possible order in which the students could have come to CTOP. If such an order does not exist, then print that this is impossible.
Please note that some students could work independently until the end of the day, without participating in a team contest.
The first line contains integer n (1 ≤ n ≤ 2·105) — the number of students who came to CTOP. The next line contains n integers a1, a2, ..., an (0 ≤ ai < n), where ai is the number of students with who the i-th student shook hands.
If the sought order of students exists, print in the first line "Possible" and in the second line print the permutation of the students' numbers defining the order in which the students entered the center. Number i that stands to the left of number j in this permutation means that the i-th student came earlier than the j-th student. If there are multiple answers, print any of them.
If the sought order of students doesn't exist, in a single line print "Impossible".
5 2 1 3 0 1
Possible 4 5 1 3 2
9 0 2 3 4 1 1 0 2 2
Possible 7 5 2 1 6 8 3 4 9
4 0 2 1 1
Impossible
In the first sample from the statement the order of events could be as follows:
- student 4 comes in (a4 = 0), he has no one to greet;
- student 5 comes in (a5 = 1), he shakes hands with student 4;
- student 1 comes in (a1 = 2), he shakes hands with two students (students 4, 5);
- student 3 comes in (a3 = 3), he shakes hands with three students (students 4, 5, 1);
- students 4, 5, 3 form a team and start writing a contest;
- student 2 comes in (a2 = 1), he shakes hands with one student (number 1).
In the second sample from the statement the order of events could be as follows:
- student 7 comes in (a7 = 0), he has nobody to greet;
- student 5 comes in (a5 = 1), he shakes hands with student 7;
- student 2 comes in (a2 = 2), he shakes hands with two students (students 7, 5);
- students 7, 5, 2 form a team and start writing a contest;
- student 1 comes in(a1 = 0), he has no one to greet (everyone is busy with the contest);
- student 6 comes in (a6 = 1), he shakes hands with student 1;
- student 8 comes in (a8 = 2), he shakes hands with two students (students 1, 6);
- student 3 comes in (a3 = 3), he shakes hands with three students (students 1, 6, 8);
- student 4 comes in (a4 = 4), he shakes hands with four students (students 1, 6, 8, 3);
- students 8, 3, 4 form a team and start writing a contest;
- student 9 comes in (a9 = 2), he shakes hands with two students (students 1, 6).
In the third sample from the statement the order of events is restored unambiguously:
- student 1 comes in (a1 = 0), he has no one to greet;
- student 3 comes in (or student 4) (a3 = a4 = 1), he shakes hands with student 1;
- student 2 comes in (a2 = 2), he shakes hands with two students (students 1, 3 (or 4));
- the remaining student 4 (or student 3), must shake one student's hand (a3 = a4 = 1) but it is impossible as there are only two scenarios: either a team formed and he doesn't greet anyone, or he greets all the three present people who work individually.
题意：
一群小盆友挨个进入教室，与教室中每个没在做contest的小盆友握手。3个小盆友可以在任意时间开始一场不会结束的contest
给出每个小盆友进教室时的握手次数。
求一个进教室的次序，满足题意。
题解：
可以发现，握手次数，增只能一个一个增，减可以幅度很大
故采取贪心策略，能增的情况就增（如果某个x在前面无法增加得到，后面就更无法达到了）
| 10808859 | 2015-04-21 13:41:24 | njczy2010 | D - Handshakes | GNU C++ | Accepted | 233 ms | 122896 KB |
1 #include <cstdio> 2 #include <cmath> 3 #include <algorithm> 4 #include <cstring> 5 #include <vector> 6 #include <queue> 7 8 using namespace std; 9 10 #define ll long long 11 12 int const N = 200005; 13 int const M = 1035; 14 ll const mod = 1000000007; 15 16 int n; 17 int a[N]; 18 19 int cnt[N]; 20 int cou[4]; 21 int ans[N]; 22 int flag; 23 queue<int> que[N]; 24 25 int judge() 26 { 27 if(n%3==0){ 28 if(cou[0]!=cou[1] || cou[0]!=cou[2] || cou[1]!=cou[2]){ 29 return 0; 30 } 31 else return 1; 32 } 33 else if(n%3==1){ 34 if(cou[0]!=cou[1]+1 || cou[0]!=cou[2]+1 || cou[1]!=cou[2]){ 35 return 0; 36 } 37 else return 1; 38 } 39 else{ 40 if(cou[0]!=cou[1] || cou[0]!=cou[2]+1 || cou[1]!=cou[2]+1){ 41 return 0; 42 } 43 else return 1; 44 } 45 return 1; 46 } 47 48 void solve() 49 { 50 int i,now; 51 for(i=0;i<=n;i++){ 52 while(que[i].size()>=1) que[i].pop(); 53 } 54 for(i=1;i<=n;i++){ 55 que[ a[i] ].push(i); 56 } 57 now=0; 58 int te; 59 for(i=1;i<=n;i++){ 60 while(now>=0){ 61 if(cnt[now]>0){ 62 te=que[now].front(); 63 que[now].pop(); 64 ans[i]=te; 65 cnt[now]--; 66 now++;break; 67 } 68 now-=3; 69 } 70 //printf(" i=%d now=%d ",i,now); 71 if(now<0){ 72 flag=0;break; 73 } 74 } 75 } 76 77 void out() 78 { 79 printf("Possible "); 80 int i; 81 printf("%d",ans[1]); 82 for(i=2;i<=n;i++){ 83 printf(" %d",ans[i]); 84 } 85 printf(" "); 86 } 87 88 int main() 89 { 90 //freopen("data.in","r",stdin); 91 //scanf("%d",&T); 92 //for(int ccnt=1;ccnt<=T;ccnt++) 93 while(scanf("%d",&n)!=EOF) 94 { 95 memset(cou,0,sizeof(cou)); 96 memset(cnt,0,sizeof(cnt)); 97 int i; 98 int j; 99 flag=1; 100 for(i=1;i<=n;i++){ 101 scanf("%d",&a[i]); 102 j=a[i]%3; 103 cnt[ a[i] ]++; 104 cou[j]++; 105 } 106 flag=judge(); 107 // printf(" flag=%d ",flag); 108 if(flag==0){ 109 printf("Impossible ");continue; 110 } 111 solve(); 112 if(flag==0){ 113 printf("Impossible ");continue; 114 } 115 out(); 116 } 117 return 0; 118 }