Generating Fast UVA10098

本文介绍如何使用C++标准模板库(STL)来高效地生成给定字符串的所有可能排列,并按照升序排序。通过实例演示了如何读取输入、排序字符串以及使用next_permutation函数生成并打印所有排列。

Problem C
Generating Fast, Sorted Permutation

Input:Standard Input

Output:Standard Output

Generating permutation has always been an important problem in computer science. In this problem you will have to generate the permutation of a given string in ascending order.Remember that your algorithm must be efficient.


Input

The first line of the input contains an integer n, which indicates how many strings to follow. The next n lines contain n strings. Strings will only contain alpha numerals and never contain any space. The maximum length of the string is 10.


Output

For each input string print all the permutations possible in ascending order. Not that the strings should be treated, as case sensitive strings and no permutation should be repeated. A blank line should follow each output set.

Sample Input

3
ab
abc
bca

Sample Output

ab
ba

abc
acb
bac
bca
cab
cba

abc
acb
bac
bca
cab
cba


和上一道题差不多,调用STL

#include<iostream>
#include<string>
#include<algorithm>

using namespace std;

bool cmp(char ch1,char ch2)
{
    return ch1<ch2;
}

int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        string str;
        cin>>str;
        sort(str.begin(),str.end(),cmp);
        int i,j;
        cout<<str<<endl;
        while(next_permutation(str.begin(), str.end()))
        {
            cout<<str<<endl;
        }
        cout<<endl;
    }
    return 0;
}






### UVA 12000: A Programming and Algorithm Perspective UVA 12000, commonly referred to in competitive programming circles, is a problem that involves algorithmic thinking and efficient implementation. While no direct description of UVA 12000 is provided in the references, we can infer potential strategies based on common patterns seen in similar problems like UVA 1363 and other Josephus problem variants. The problem likely involves a sequence or array manipulation, possibly with modular arithmetic or cyclic behavior. Based on the pattern of other UVA problems involving recursive or iterative logic, UVA 12000 may require a solution that leverages mathematical observations to reduce complexity. For instance, if UVA 12000 involves a similar structure to the Josephus problem, the solution could be expressed using a loop that iteratively computes the position of the survivor or some other derived value. The following code snippet demonstrates a general approach that could be adapted depending on the exact requirements of the problem: ```python def solve(n, k): r = 0 for i in range(1, n + 1): r = (r + k) % i return r ``` This function computes a result using an iterative approach where `k` is added to the current result `r` and then the modulo operation is applied with the current index `i`. This is a common technique in problems involving circular elimination or selection processes. In the case of UVA 12000, the problem may require additional constraints or a variation of this logic. For example, it could involve handling large input sizes efficiently, which would necessitate optimizing the algorithm to avoid unnecessary computations. Techniques such as memoization or dynamic programming could play a role in achieving this efficiency. When implementing a solution, it is crucial to consider edge cases, such as when `n` or `k` is very small or when `k` exceeds `n`. Handling these cases correctly ensures robustness in the implementation. For example, if `k` is larger than `i` during the loop iteration, the modulo operation naturally reduces it to an equivalent smaller value. If the problem involves prime numbers or paths, such as in UVA 12101, then a breadth-first search (BFS) or similar graph traversal technique might be required. In such cases, the solution would involve generating valid transitions between states and efficiently exploring the search space. ###
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