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  • PAT 1067. Sort with Swap(0,*)

    Given any permutation of the numbers {0, 1, 2,..., N-1}, 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 (<=105) followed by a permutation sequence of {0, 1, ..., N-1}. 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

    分析
    这道题用到了表排序中N个数字的排列由若干独立的环组成的知识。
    1.单元环swap(0,i)次数为0;
    2.有0的非单元环,swap(0,i)次数为n-1;
    3.无0的单元环,可以先把环里随便一个数和0交换一次,这样整个环就变成了含0的n+1个元素的非单元环,根据前面的情况2,次数为(n+1)-1,但还要加上把0换进去的那次,故swap(0,i)次数为n+1;
    代码如下:

    #include<iostream>
    using namespace std;
    int main(){
    	int n,sum=0;
    	cin>>n;
    	int a[n],visited[n]={0};
    	for(int i=0;i<n;i++)
    	    cin>>a[i];
    	for(int i=0;i<n;i++){
    		int temp=i,cnt=0,flag=0;
    		while(visited[temp]!=1){
    			 cnt++;
    			 if(a[temp]==0) flag=1;
    			 visited[temp]=1;
    			 temp=a[temp];
    		}
    		if(cnt>1&&flag==1) sum+=cnt-1;
    		else if(cnt>1&&flag==0) sum+=cnt+1;
    	}
    	cout<<sum;
    	return 0;
    } 
    
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  • 原文地址:https://www.cnblogs.com/A-Little-Nut/p/8371333.html
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