Problem UVA11925-Generating Permutations
Accept: 214 Submit: 1429
Time Limit: 1000 mSec
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Problem Description
A permutation on the integers from 1 to n is, simply put, a particular rearrangement of these integers. Your task is to generate a given permutation from the initial arrangement 1,2,3,...,n using only two simple operations.
? Operation 1: You may swap the ?rst two numbers. For example, this would change the arrangement 3,2,4,5,1 to 2,3,4,5,1.
? Operation 2: You may move the ?rst number to the end of the arrangement. For example, this would change the arrangement 3,2,4,5,1 to 2,4,5,1,3.
Input
The input consists of a number of test cases. Each test case begins with a single integer n between 1 and 300. On the same line, a permutation of integers 1 through n is given where consecutive integers are separated by a single space. Input is terminated by a line containing ‘0’ which should not be processed.
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Output
For each test case you are to output a string on a single line that describes a sequence of operations. The string itself should consist only of the characters ‘1’ and ‘2’. This string should be such that if we start with the initial arrangement 1,2,3,...,n?1,n and successively apply rules 1 and 2 according to the order they appear in the output, then the resulting permutation is identical to the input permutation. The output string does not necessarily need to be the shortest such string, but it must be no longer than 2n2 characters. If it is
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Sample Input
3 2 1 3
3 2 3 1
4 4 2 3 1
0
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Sample Output
1
2
12122
题解：这个题首先应该转换一下思维，考虑将给定串排成升序而不是将排好序的串变成给定串，这样会好想很多，注意如果这样思考的话，1操作就变成把最后一个数移到最前面，2操作不受影响。排序就是一个消除逆序对的过程，所以如果前面两个数是满足第一个数大于第二个数，那就要通过交换来消除这个逆序对（这样操作次数少），这里有个特殊情况就是第一个数是n并且第二个数是1，这时虽然构成逆序，但是是有可能通过把后面的数移到前面而使序列有序的，所以这时不要交换。
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 int n;
6 deque<int> seq;
7 string ans;
8
9 bool check() {
10 for (int i = 0; i < n; i++) {
11 if (seq[i] != i + 1) return false;
12 }
13 return true;
14 }
15
16 int main()
17 {
18 //freopen("input.txt", "r", stdin);
19 while (~scanf("%d", &n) && n) {
20 seq.clear();
21 ans = "";
22 int x;
23 for (int i = 0; i < n; i++) {
24 scanf("%d", &x);
25 seq.push_back(x);
26 }
27
28 while (true) {
29 if (seq[0] == 1 && check()) {
30 break;
31 }
32 if (seq[0] < seq[1] || seq[0] == n && seq[1] == 1) {
33 seq.push_front(seq[n - 1]);
34 seq.pop_back();
35 ans += ‘2‘;
36 }
37 else {
38 swap(seq[0], seq[1]);
39 ans += ‘1‘;
40 }
41 }
42 reverse(ans.begin(), ans.end());
43 cout << ans << endl;
44 }
45 return 0;
46 }
原文地址：https://www.cnblogs.com/npugen/p/9685888.html