问题 F: Escape Room

时间限制: 1 Sec 内存限制: 128 MB
提交: 63 解决: 27
[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


**************************************************************************
是一道关于最长上升子序列的思维题吧 就归到了dp里
题意：
给你n个数   是每个数从该位置到末位的最长上升子序列的最大长度 
求一个符合的数列（如果多个输出最小）
//
例如样例输出 ：4   2    1    3   
（4的最长上升长度为1；2的最长上升长度为2；1的最长上升长度为2；3的最长上升长度为1）
即为样例输入：1   2    2    1

一开始的想法就是 1   2   2   1每次遍历最小的将4 3 2 1 顺序填入
但是每次都要遍历，我就试着同时遍历最大和最小，但还是tle了
我的tle错误代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
 
using namespace std;
 
int main()
{
    int n;
    int a[100005]={0};
    int ans[100005]={0};
    scanf("%d",&n);
    int maxx = 0;
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        if(a[i]>maxx)
            maxx = a[i];
    }
    int p=1,q=maxx; //min max order
    int cnt1=1,cnt2=n; // min max num
    while(1)
    {
        if(p>q)
            break;
        for(int i=0;i<n;i++)
        {
            if(a[i]==p)
            {
                ans[i]=cnt2;
                cnt2--;
            }
            if(a[n-1-i]==q)
            {
                ans[n-1-i]=cnt1;
                cnt1++;
            }
        }
//        for(int i=0;i<n;i++)
//        {
//            printf("%d ",ans[i]);
//        }
        //printf("\n");
        p++;
        q--;
    }
    for(int i=0;i<n-1;i++)
    {
        printf("%d ",ans[i]);
    }
    printf("%d",ans[n-1]);
}

然后队友想了一种算法：用三维的结构体

为了更容易明白我写了组样例：1 1 2 3 3 2 1 1

第一步：将第一维的最大上升序列的下标记在第二维（用于记录原来的位置）：

第一维x: 1 1 2 3 3 2 1 1

第二维num:1 2 3 4 5 6 7 8

按第一维排序：

x: 1 1 1 1 2 2 3 3

num：1 2 7 8 3 6 4 5

第三维：8 7 6 5 4 3 2 1（倒序填入对应的x）

为了将x变回原本的位置

将整个结构体按num再次排序

返回原本的顺序

x： 1 1 2 3 3 2 1 1

num：1 2 3 4 5 6 7 8

r： 8 7 4 2 1 3 6 5

输出的第三维就是答案

正确代码：

#include <sstream>
#include <iostream>
#include <string>
#include<stdlib.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
struct po    //结构体
{
    int x,num,r;
};
bool cmp1(po a,po b)  
{
    if(a.x!=b.x)
        return a.x<b.x;
    return a.num<b.num;

}
bool cmp2(po a,po b)
{
    return a.num<b.num;
}
int main()
{
    struct po s[100008]= {0};
    int n,i,x,t;
    scanf("%d",&n);
    for(i=1; i<=n; i++)
    {
        scanf("%d",&t);
        s[i].num=i;  //将原本顺序填入num中
        s[i].x=t;    
    }
    sort(s+1,s+1+n,cmp1);  //按x排
    int k=n;
    for(i=1; i<=n; i++)    //倒序填入
    {
        s[i].r=k;
        k--;
    }
    sort(s+1,s+1+n,cmp2);  //按num排序 使结构体恢复原本顺序
    for(i=1;i<n;i++)printf("%d ",s[i].r);  //输出 r 
    printf("%d\n",s[n].r);
    return 0;
}