B. Minimize the Permutation (贪心)

时间:2022-07-28
本文章向大家介绍B. Minimize the Permutation (贪心),主要内容包括其使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。

题意描述

You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

You can perform at most n−1 operations with the given permutation (it is possible that you don’t perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.

Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.

You can see the definition of the lexicographical order in the notes section.

You have to answer q independent test cases.

For example, let’s consider the permutation [5,4,1,3,2]. The minimum possible permutation we can obtain is [1,5,2,4,3] and we can do it in the following way:

perform the second operation (swap the second and the third elements) and obtain the permutation [5,1,4,3,2]; perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5,1,4,2,3]; perform the third operation (swap the third and the fourth elements) and obtain the permutation [5,1,2,4,3]. perform the first operation (swap the first and the second elements) and obtain the permutation [1,5,2,4,3]; Another example is [1,2,4,3]. The minimum possible permutation we can obtain is [1,2,3,4] by performing the third operation (swap the third and the fourth elements).

Input The first line of the input contains one integer q (1≤q≤100) — the number of test cases. Then q test cases follow.

The first line of the test case contains one integer n (1≤n≤100) — the number of elements in the permutation.

The second line of the test case contains n distinct integers from 1 to n — the given permutation.

Output For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.

Example input 4 5 5 4 1 3 2 4 1 2 4 3 1 1 4 4 3 2 1 output 1 5 2 4 3 1 2 3 4 1 1 4 3 2

思路

记录下每个数字的初始位置,然后尽量让每个数都回到和该数字相同的位置上,比如让数字1回到下标1的位置

AC代码

#include<bits/stdc++.h>
#define x first
#define y second
#pragma GCC optimize(2)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize(3)
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC target("sse3","sse2","sse")
#pragma GCC target("avx","sse4","sse4.1","sse4.2","ssse3")
#pragma GCC target("f16c")
#pragma GCC optimize("inline","fast-math","unroll-loops","no-stack-protector")
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"
#pragma GCC diagnostic error "-std=c++14"
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef long long LL;
const int N=105;
const int M=150;
const int INF=0x3f3f3f3f;
const int MOD=998244353;
int a[N];
bool st[N];
int pos[N];
int main(){
    IOS;
    int T;cin>>T;
    while(T--){
        memset(st,false,sizeof st);
        memset(pos,0,sizeof pos);
        memset(a,0,sizeof a);
        int n;cin>>n;
        for(int i=1;i<=n;i++){
            cin>>a[i];
            pos[a[i]]=i;
        }
        int t=n-1;
        int idx=1;
        for(int i=1;i<=n;i++){
            int flag=1;
            while(flag){
                if(pos[i]!=i&&!st[pos[i]-1]){
                    st[pos[i]-1]=true;
                    swap(a[pos[i]],a[pos[i]-1]);
                    swap(pos[a[pos[i]]],pos[a[pos[i]-1]]);
                }else{
                    flag=0;
                }
            }
            st[pos[i]]=true;
        }
        for(int i=1;i<=n;i++) cout<<a[i]<<' ';
        cout<<endl;
    }
    return 0;
}