题目链接：HDU 5775

题面：

Bubble Sort

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

P is a permutation of the integers from 1 to N(index starting from 1).

Here is the code of Bubble Sort in C++.

for(int i=1;i<=N;++i)
    for(int j=N,t;j>i;—j)
        if(P[j-1] > P[j])
            t=P[j],P[j]=P[j-1],P[j-1]=t;

After the sort, the array is in increasing order. ?? wants to know the absolute values of difference of rightmost place and leftmost place for every number it reached.

Input

The first line of the input gives the number of test cases T; T test cases follow.

Each consists of one line with one integer N, followed by another line with a permutation of the integers from 1 to N, inclusive.

limits

T <= 20

1 <= N <= 100000

N is larger than 10000 in only one case.

Output

For each test case output “Case #x: y1 y2 … yN” (without quotes), where x is the test case number (starting from 1), and yi is the difference of rightmost place and leftmost place of number i.

Sample Input

2
3
3 1 2
3
1 2 3

Sample Output

Case #1: 1 1 2
Case #2: 0 0 0

Hint

In first case, (3, 1, 2) -> (3, 1, 2) -> (1, 3, 2) -> (1, 2, 3)
the leftmost place and rightmost place of 1 is 1 and 2, 2 is 2 and 3, 3 is 1 and 3
In second case, the array has already in increasing order. So the answer of every number is 0.


 

Author

FZU

Source

2016 Multi-University Training Contest 4

题意：

给定一个1-N的随机排列，对此序列进行冒泡排序，问在排序过程中，每个数到达过的最右端和最左端的差值。

解题：

因为排序的具体过程是从右侧开始，每次将当前最小的值移到最左端，故一个数会到达的最右位置是它的起始位置+其右侧小于它的数字数量（在后面比他小的数向左移动的过程中，它会向右移动这么多步）。最左位置是其原来位置和排序之后应该处于的位置，两者的小者。最后，两个位置作差即为所求。

关键点是在于如何统计一个数右侧有几个数比它小，可以从右往左遍历，用树状数组维护，因为是边移动边统计的，每次就可以直接询问【1-a[j]-1】的和，即为右侧小于a[j]的数字的数量。

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#define inf 1000000
using namespace std;
int a[100010],c[100010],ans[100010];
int t,n;
int max(int a,int b)
{
	return a>b?a:b;
}
int min(int a,int b)
{
	return a<b?a:b;
}
int lowbit(int x)
{
	return x&(-x);
}
int sum(int x)
{
	int res=0;
	while(x>0)
	{
		res+=c[x];
		x-=lowbit(x);
	}
	return res;
}
void update(int i,int val)
{
	while(i<=n)
	{
		c[i]+=val;
		i+=lowbit(i);
	}
}
int main()
{
	int tmp,le,ri;
	scanf("%d",&t);
    for(int i=1;i<=t;i++)
	{
		scanf("%d",&n);
		for(int j=0;j<n;j++)
           scanf("%d",&a[j]);
		memset(c,0,sizeof(c));
        for(int j=n-1;j>=0;j--)
		{
           tmp=sum(a[j]);
		   update(a[j],1);
		   le=min(a[j]-1,j);
		   ri=j+tmp;
		   ans[a[j]]=ri-le;
		}
		printf("Case #%d:",i);
		for(int j=1;j<=n;j++)
			printf(" %d",ans[j]);
		printf("\n");
	}
	return 0;
}