HDU 5775 Bubble Sort

Bubble Sort

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

Total Submission(s): 827 Accepted Submission(s): 493

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.

题意：题意：给你一个n的排列，然后对它做一遍冒泡排序，问每个数在排序中的最左边位置和最右边位置的差是多少。

思路：分析：对于一个数i,若它的右边还有比它还小的数，那么它在排序过程中还会向右移动，否则向左；当i在的位置小于i时，答案为i右边比它小的数的个数，当i的位置大于等于i时，答案为最后位置减去初始位置加上右边比它小的数，后者可以用树状数组求出。

代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
#define Max 155555
int node[Max<<2],m[Max],ans[Max],x[Max];
void build(int l,int r,int rt)
{
node[rt]=0;
if(l==r)
{
return;
}
int m=(l+r)>>1;
build(lson);
build(rson);
}
void modify(int p,int l,int r,int rt)
{
if(l==r)
{
node[rt]=1;
return;
}
int m=(l+r)>>1;
if(p<=m)
{
modify(p,lson);
}
else
{
modify(p,rson);
}
node[rt]=node[rt<<1]+node[rt<<1|1];
}
int query(int L,int R,int l,int r,int rt)
{
if(L<=l&&R>=r)
{
return node[rt];
}
int m=(l+r)>>1;
int sum=0;
if(L<=m)
{
sum+=query(L,R,lson);
}
if(R>m)
{
sum+=query(L,R,rson);
}
return sum;
}
int main()
{
int t,t1=0;
scanf("%d",&t);
while(t--)
{
int n,i;
t1++;
//memset(ans,0,sizeof(ans));
scanf("%d",&n);
build(1,n,1);
for(i=1;i<=n;i++)
{
scanf("%d",&m[i]);
x[m[i]]=i;
}
for(i=n;i>=1;i--)
{
ans[m[i]]=query(1,m[i],1,n,1);
modify(m[i],1,n,1);
}
printf("Case #%d:",t1);
for(i=1;i<=n;i++)
{
if(x[i]<=i)
printf(" %d",ans[i]);
else
{
printf(" %d",ans[i]+abs(x[i]-i));
}
}
printf("\n");
}
return 0;
}