5140 Escape Room SEERC2017

http://exam.upc.edu.cn/problem.php?cid=1292&pid=5

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


题目大意：输入一组数据Ai，Ai表示以下标i为起点的最大递增序列的长度为Ai，求满足输入数据的一组序列P，且保证这组序列按字典排序是最小的。
解决方法：我们把Ai与Pi一一对应，然后把Ai按从小到大排列，同时Pi也跟着排列，我们发现最终的结果满足这种处理后的Pi是从大到小排列的，而Ai是从小到大排列的
　　　　　在实现这个代码时，一开始我们就想到用sort函数对Ai进行排序，但是在写比较条件的时候，我们忘了判断当Ai中有两个相等的值的情况。主要原因在于对sort函数的不理解，
  　　　　而我们试的样例范围很小，所以我们试的都过了，但是提交后就WA

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
struct node
{
    int val;
    int pos;
} lin[100050];
int ans[100050];
bool cmp(node a,node b)
{
    if(a.val==b.val)
    {
        return a.pos<b.pos;
    }
    return a.val<b.val;
}
int main()
{
    int n,m,i;
    scanf("%d",&n);
    m=n;
    for(i=0; i<n; i++)
    {
        scanf("%d",&lin[i].val);
        lin[i].pos=i;
    }
    sort(lin,lin+n,cmp);
    for(i=0; i<n; i++)
    {
        ans[lin[i].pos]=m;
        m--;
    }
    for(i=0; i<n-1; i++)
    {
        printf("%d ",ans[i]);
    }
    printf("%d\n",ans[n-1]);
    return 0;
}