CodeForces 471B MUH and Important Things 要求输出

原题： http://codeforces.com/contest/471/problem/B

题目：

MUH and Important Things

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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

4

1 3 3 1

output

YES

1 4 2 3

4 1 2 3

4 1 3 2

input

5

2 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.

思路：

输入每个数的优先级梯度，输出可能出现的情况。

显然，只要有三个数在同一优先级或者有两组两个数在同一优先级就可以输出结果。

代码：

#include <iostream>
#include"string.h"
#include"cstdio"
#include"stdlib.h"
#include"algorithm"
#include"queue"
#include"stack"

using namespace std;
typedef long long int lint;
#define N 2010

typedef struct
{
int xuhao;
int dj;
} jgt;

int main()
{
int n;
jgt st
;

while(scanf("%d",&n)!=EOF)
{
queue <jgt> q;
memset(st,0,sizeof(st));
for(int i=0; i<n; i++)
{
st[i].xuhao=i+1;
scanf("%d",&st[i].dj);
}

int tem=0;
for(int i=1; i<N; i++)
{
for(int j=0; j<n; j++)
{
if(st[j].dj==i)
{
q.push(st[j]);
tem++;
}
}
if(tem==n)
{
break;
}
}

int a
;    //dj
int b
;    //xuhao
int c[2];
for(int i=0; i<n; i++)
{
a[i]=q.front().dj;
b[i]=q.front().xuhao;
q.pop();
}
int j=0;
for(int i=0; i<n-1; i++)
{
if(a[i]==a[i+1])    c[j++]=i;
if(j==2)    break;
}
if(j<2) printf("NO\n");
else
{
printf("YES\n");
for(int i=0; i<n; i++)
{
printf("%d ",b[i]);
}
printf("\n");
swap(b[c[0]],b[c[0]+1]);
for(int i=0; i<n; i++)
{
printf("%d ",b[i]);
}
printf("\n");
swap(b[c[1]],b[c[1]+1]);
for(int i=0; i<n; i++)
{
printf("%d ",b[i]);
}
printf("\n");
}
}
return 0;
}