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  • 10-排序6 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
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    const int maxn = 100010;
    int pos[maxn];
    int main(){
        int n,ans = 0;
        scanf("%d",&n);
        int left = n - 1,num;
        for(int i = 0; i < n; i++){
            scanf("%d",&num);
            pos[num] = i;
            if(num == i && num != 0){
                left--;
            }
        }
        int k = 1;
        while(left > 0){
            if(pos[0] == 0){
                while(k < n){
                    if(pos[k] != k){
                        swap(pos[0],pos[k]);
                        ans++;
                        break;
                    }
                    k++;
                }
            } 
            if(pos[0] != 0){
                swap(pos[0],pos[pos[0]]);
                ans++;
                left--; 
            }
        }
        printf("%d",ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/wanghao-boke/p/10409287.html
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