标签：mina art strong data 技术 result tween should nta
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.
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.
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 }
UVA11925-Generating Permutations（贪心）
标签：mina art strong data 技术 result tween should nta
原文地址：https://www.cnblogs.com/npugen/p/9685888.html
