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  • PAT Advanced 1067 Sort with Swap(0, i) (25分)

    Given any permutation of the numbers {0, 1, 2,..., N1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may apply the swap operations in the following way:

    Swap(0, 1) => {4, 1, 2, 0, 3}
    Swap(0, 3) => {4, 1, 2, 3, 0}
    Swap(0, 4) => {0, 1, 2, 3, 4}
    
     

    Now you are asked to find the minimum number of swaps need to sort the given permutation of the first N nonnegative integers.

    Input Specification:

    Each input file contains one test case, which gives a positive N (≤) followed by a permutation sequence of {0, 1, ..., N1}. All the numbers in a line are separated by a space.

    Output Specification:

    For each case, simply print in a line the minimum number of swaps need to sort the given permutation.

    Sample Input:

    10
    3 5 7 2 6 4 9 0 8 1
    
     

    Sample Output:

    9
    
     

    依旧采用柳神的办法(好精巧,不知道怎么想到的),柳太强了。

    0号哨兵。第一位开始遍历,如果不与该位置相等,0号一直调整到相应位置,自己和0号位置交换。

    #include <iostream>
    #include <map>
    #include <algorithm>
    using namespace std;
    int main() {
        int N, cnt = 0, tmp;
        scanf("%d", &N);
        map<int, int> m;
        for(int i = 0; i < N; i++){
            scanf("%d", &tmp);
            m[tmp] = i;
        }
        for(int i = 1; i < N; i++) {
            if(i != m[i]) {
                while(m[0] != 0) {
                    swap(m[0], m[m[0]]);
                    cnt++;
                }
                if(i != m[i]) {
                    swap(m[0], m[i]);
                    cnt++;
                }
            }
        }
        cout << cnt;
        system("pause");
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/littlepage/p/12268544.html
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