CF#269 (Div. 2) B

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It's time polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got down to business. In total, there are n tasks for the day and each animal should do each of these tasks. For each task, they have evaluated its difficulty. Also animals decided to do the tasks in order of their difficulty. Unfortunately, some tasks can have the same difficulty, so the order in which one can perform the tasks may vary.

Menshykov, Uslada and Horace ask you to deal with this nuisance and come up with individual plans for each of them. The plan is a sequence describing the order in which an animal should do all the n tasks. Besides, each of them wants to have its own unique plan. Therefore three plans must form three different sequences. You are to find the required plans, or otherwise deliver the sad news to them by stating that it is impossible to come up with three distinct plans for the given tasks.

Input

The first line contains integer n (1 ≤ n ≤ 2000) — the number of tasks. The second line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 2000), where hi is the difficulty of the i-th task. The larger number hi is, the more difficult the i-th task is.

Output

In the first line print "YES" (without the quotes), if it is possible to come up with three distinct plans of doing the tasks. Otherwise print in the first line "NO" (without the quotes). If three desired plans do exist, print in the second line n distinct integers that represent the numbers of the tasks in the order they are done according to the first plan. In the third and fourth line print two remaining plans in the same form.

If there are multiple possible answers, you can print any of them.

Sample test(s)
input
41 3 3 1
output
YES1 4 2 3 4 1 2 3 4 1 3 2 
input
52 4 1 4 8
output
NO

Note

In the first sample the difficulty of the tasks sets one limit: tasks 1 and 4 must be done before tasks 2 and 3. That gives the total of four possible sequences of doing tasks : [1, 4, 2, 3], [4, 1, 2, 3], [1, 4, 3, 2], [4, 1, 3, 2]. You can print any three of them in the answer.

In the second sample there are only two sequences of tasks that meet the conditions — [3, 1, 2, 4, 5] and [3, 1, 4, 2, 5]. Consequently, it is impossible to make three distinct sequences of tasks.

解题思路：

题目大意是问n个数按权值从小到大排序，能否得到三个序列号不一样的数组，能输出YES和任意三个符合要求的数组，否则输出NO。

趁中午吃饭的功夫，借鉴了下sxk大牛的思想，排完序后扫的时候只要找到两对相等的就停止扫描（即cnt == 3），当YES时，第一次输出排完序的结果；第二次正着扫把第一对调换输出；第三次逆着扫把最后一对调换输出。

完整代码：

#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")typedef long long LL;typedef double DB;typedef unsigned uint;typedef unsigned long long uLL;/** Constant List .. **/ //{const int MOD = int(1e9)+7;const int INF = 0x3f3f3f3f;const LL INFF = 0x3f3f3f3f3f3f3f3fLL;const DB EPS = 1e-9;const DB OO = 1e20;const DB PI = acos(-1.0); //M_PI;struct node{    int num;    int h;}q[3010];int n;bool cmp(node a , node b){    if(a.h == b.h)        return a.num < b.num;    else        return a.h < b.h;}void display(){    for(int i = 0 ; i < n ; i ++)    {        printf("%d%c",q[i].num , i == n - 1 ? '\n' : ' ');    }}void one(){    int t;    for(int i = 0 ; i < n ; i ++)    {        if(q[i].h == q[i+1].h && i + 1 < n)        {                t = q[i].num;                q[i].num = q[i+1].num;                q[i+1].num = t;                break;        }    }    display();}void two(){    int t;    for(int i = n - 1 ; i >= 0 ; i --)    {        if(q[i].h == q[i-1].h && i - 1 >= 0)        {                t = q[i].num;                q[i].num = q[i-1].num;                q[i-1].num = t;                break;        }    }    display();}int main(){    #ifdef DoubleQ    freopen("in.txt","r",stdin);    #endif    while(~scanf("%d",&n))    {        for(int i = 0 ; i < n ; i++)        {            scanf("%d",&q[i].h);            q[i].num = i + 1;        }        sort(q , q + n , cmp);        int cnt = 1;        for(int i = 0 ; i < n ; i ++)        {            if(q[i].h == q[i+1].h && i + 1 < n)            {                cnt ++;            }            if(cnt == 3)                break;        }        if(cnt < 3)        {            printf("NO\n");        }        else        {            printf("YES\n");            display();            one();            two();        }    }}

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