B. MUH and Important Things（Codeforces Round #269)

最新推荐文章于 2019-01-22 00:20:43 发布

原创最新推荐文章于 2019-01-22 00:20:43 发布 · 1.1k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#algorithm

Codeforces 专栏收录该内容

61 篇文章

订阅专栏

探讨了为三只动物设计独特任务执行计划的问题，确保每只动物的任务执行顺序不同且符合任务难度约束。

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.

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
vector<int> son[2014];
int main()
{
    int n;
    scanf("%d",&n);
    int x;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&x);
        son[x].push_back(i);
    }
    int coun=0;
    int sign=0;
    for(int i=1;i<=2000;i++)
    {
        if(son[i].size()==2)
        {
            coun+=2;
        }
        if(son[i].size()>2)
        {
            sign=i;
            coun+=4;
            break;
        }
    }
    if(coun>=3)
    {
        printf("YES\n");
        int cr=1;
        for(int i=1;i<=2000;i++)
        {
            for(int j=0;j<son[i].size();j++)
            {
                if(cr)
                {
                    cr=0;
                    printf("%d",son[i][j]);
                }
               else
                   printf(" %d",son[i][j]);
            }
        }
        printf("\n");
        if(sign)
        {
            cr=1;
            for(int i=1;i<=2000;i++)
            {
            for(int j=0;j<son[i].size();j++)
            {
                if(i==sign&&(j==0||j==1))
                {
                    if(cr)
                   {
                    cr=0;
                    printf("%d",son[i][1-j]);
                   }
                   else
                    printf(" %d",son[i][1-j]);
                }
                else
                {
                if(cr)
                {
                    cr=0;
                    printf("%d",son[i][j]);
                }
               else
                    printf(" %d",son[i][j]);
                }
            }
           }
           printf("\n");
           cr=1;
            for(int i=1;i<=2000;i++)
            {
            for(int j=0;j<son[i].size();j++)
            {
                if(i==sign&&(j==0||j==1||j==2))
                {
                    if(cr)
                   {
                    cr=0;
                    printf("%d",son[i][2-j]);
                   }
                   else
                    printf(" %d",son[i][2-j]);
                }
                else
                {
                if(cr)
                {
                    cr=0;
                    printf("%d",son[i][j]);
                }
               else
                    printf(" %d",son[i][j]);
                }
            }
           }
           printf("\n");
        }
        else
        {
            int gt=0;
            cr=1;
            for(int i=1;i<=2000;i++)
            {
            for(int j=0;j<son[i].size();j++)
            {
                if(son[i].size()==2&&(j==0||j==1))
                {
                    gt=i;
                    if(cr)
                   {
                    cr=0;
                    printf("%d",son[i][1-j]);
                   }
                   else
                    printf(" %d",son[i][1-j]);
                }
                else
                {
                if(cr)
                {
                    cr=0;
                    printf("%d",son[i][j]);
                }
               else
                    printf(" %d",son[i][j]);
                }
            }
           }
           printf("\n");
           cr=1;
            for(int i=1;i<=2000;i++)
            {
            for(int j=0;j<son[i].size();j++)
            {
                if(son[i].size()==2&&(j==0||j==1)&&i!=gt)
                {

                    if(cr)
                   {
                    cr=0;
                    printf("%d",son[i][1-j]);
                   }
                   else
                    printf(" %d",son[i][1-j]);
                }
                else
                {
                if(cr)
                {
                    cr=0;
                    printf("%d",son[i][j]);
                }
               else
                    printf(" %d",son[i][j]);
                }
            }
           }
           printf("\n");
        }
    }
    else
    {
        printf("NO\n");
    }
    return 0;
}