哈理工oj 1372 Next Permutation【STL真好用系列】

最新推荐文章于 2017-01-21 23:48:31 发布

原创最新推荐文章于 2017-01-21 23:48:31 发布 · 481 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#hrbust 1372 #哈理工oj 1372

水题专栏收录该内容

175 篇文章

订阅专栏

本文探讨了如何在给定的排列中找到下一个更高的排列，适用于计算机科学领域内的排序算法研究。

Next Permutation

Time Limit: 1000 MS

Memory Limit: 65536 K

Total Submit: 98(49 users)

Total Accepted: 60(46 users)

Rating:

Special Judge: No

Description

A permutation is each one of the (n!) possible arrangements the elements in a set can take (where n is the number of elements in the set). Different permutations can be ordered according on how they compare lexicographically
 to each other; the first such-sorted possible permutation (the one that would compare lexicographically smaller to all other permutations) is the one which has all its elements sorted in ascending order, and the largest has all its elements sorted in descending
 order.
You need to determine the next k-th higher permutation of a given permutation of the set {1, 2, 3 … n}.

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases. Then T test cases follow.
For each test case:
Line 1. This line contains two space-separated integers n and k (1 ≤ n, k ≤ 1000)
Line 2. This line contains n space-separated integers indicating a permutation of {1, 2, 3 …n}

Output

For each test case:
Line 1. Output the next k-th higher permutation of the set {1, 2, 3 … n}.Separate the numbers by a single space.

Sample Input

3
1
3 1
1
2 1
2
2 3 4 5 6 7 8 9 10

Sample Output

1 2
2 3
2 3 4 5 6 7 9 8 10

Hint

In the first sample, all the permutations of {1, 2, 3} in lexicographical order are as below:
{1, 2, 3}
{1, 3, 2}
{2, 1, 3}
{2, 3, 1}
{3, 1, 2}
{3, 2, 1}
So the next first permutation of {2, 3, 1} is {3, 1, 2}.

Source

哈理工2012春季校赛 - 网络预选赛

Author

齐达拉图@HRBUST

如果有兴趣的小伙伴可以去做做hdu 1027 这两个题是一样的，只不过这个的输入稍微复杂些，直接套用库函数就行了~：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[10000];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=m;i++)
        {
            next_permutation(a+1,a+n+1);
        }
        printf("%d",a[1]);
        for(int i=2;i<=n;i++)
        {
            printf(" %d",a[i]);
        }
        printf("\n");
    }
}