H - Frosh Week

最新推荐文章于 2019-02-08 19:38:00 发布

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最新推荐文章于 2019-02-08 19:38:00 发布

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原文链接：http://www.cnblogs.com/lbyj/p/5696919.html

本文介绍了一种使用归并排序算法来解决逆序对计数问题的方法。通过将学生队伍按学号排序的过程，实现了逆序对的有效计数。文章详细解释了归并排序的实现过程，并给出了完整的AC代码。

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Description

During Frosh Week, students play various fun games to get to know each other and compete against other teams. In one such game, all the frosh on a team stand in a line, and are then asked to arrange themselves according to some criterion, such as their height, their birth date, or their student number. This rearrangement of the line must be accomplished only by successively swapping pairs of consecutive students. The team that finishes fastest wins. Thus, in order to win, you would like to minimize the number of swaps required.

Input

The first line of input contains one positive integer n, the number of students on the team, which will be no more than one million. The following n lines each contain one integer, the student number of each student on the team. No student number will appear more than once.

Output

Output a line containing the minimum number of swaps required to arrange the students in increasing order by student number.

Sample Input

Sample Output

AC代码：

#include<iostream>//归并排序
#include<cstdio>
#include<algorithm>
#include<cstring>
# define M 1000010
using namespace std;
long long int cnt;
int A[M],T[M];
int y,m;
void merge_sort(int*A,int x,int y,int *T)
{
  if(y-x>1)
  {
   int m=x+(y-x)/2;//划分
   int p=x,q=m,i=x;
   merge_sort(A,x,m,T);//递归求解
   merge_sort(A,m,y,T);//递归求解
   while(p<m||q<y)
   {
     if(q>=y||(p<m&&A[p]<=A[q]) )T[i++]=A[p++];//从左半数组复制到临时空间
	 else
	 {
		 T[i++]=A[q++];//从右半数组复制到临时空间
		 cnt+=m-p;
	 }
   }
	 for(i=x;i<y;i++)
		 A[i]=T[i];//从辅助空间复制回A数组
  }
}

int main()
{
	int n;

	while(scanf("%d",&n)!=EOF)
	{
	    cnt=0;
		memset(A,0,sizeof(A));
		memset(T,0,sizeof(T));
		int i=0;
		while(n--)
		{
	scanf("%d",&A[i++]);
		}
    merge_sort(A,0,i,T);
		printf("%lld\n",cnt);

	}

return 0;

}