Escape Room

5140: Escape Room

时间限制: 1 Sec   内存限制: 128 MB
提交: 116   解决: 53
[ 提交][ 状态][ 讨论版][命题人: admin]

题目描述

As you know, escape rooms became very popular since they allow you to play the role of a video game hero. One such room has the following quiz. You know that the locker password is a permutation of N numbers. A permutation of length N is a sequence of distinct positive integers, whose values are at most N. You got the following hint regarding the password - the length of the longest increasing subsequence starting at position i equals Ai. Therefore you want to find the password using these values. As there can be several possible permutations you want to find the lexicographically smallest one. Permutation P is lexicographically smaller than permutation Q if there is an index i such that Pi < Qi and Pj = Qj for all j < i. It is guaranteed that there is at least one possible permutation satisfying the above constraints. 
Can you open the door?

输入

The first line of the input contains one integer N (1≤N≤10⁵).
The next line contains N space-separated integer Ai (1≤Ai≤N).
It’s guaranteed that at least one possible permutation exists.

输出

Print in one line the lexicographically smallest permutation that satisfies all the conditions.

样例输入

4
1 2 2 1

样例输出

4 2 1 3

题意：给你一个数n和n个数ai，ai表示从i开始到第n个数的最长上升子序列的个数，输出满足条件的一组且字典序最小的一组数

思路：按ai从小到大排序，ai相同的按输入顺序排序，从n开始按顺序赋值，变回原来的顺序，输出

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;
typedef struct node
{
    int x,index,y;
}NODE;
NODE p[100010];
int cmp1(NODE a,NODE b)
{
    if(a.x!=b.x)
        return a.x<b.x;
    return a.index<b.index;
}
int cmp2(NODE a,NODE b)
{
    return a.index<b.index;
}
int main()
{
    int n,i,j;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&p[i].x);
        p[i].index=i;
    }
    sort(p+1,p+n+1,cmp1);
    for(i=1;i<=n;i++)
        p[i].y=(n-i+1);
    sort(p+1,p+n+1,cmp2);
    printf("%d",p[1].y);
    for(i=2;i<=n;i++)
        printf(" %d",p[i].y);
    putchar('\n');
    return 0;
}