多校训练-Bubble Sort（树状数组+离散化）

Bubble Sort

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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

233 1 231 2 3

 

Sample Output

Case #1: 1 1 2Case #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 3In second case, the array has already in increasing order. So the answer of every number is 0.

 

大致题意：给你一串数，求其在冒泡排序过程中任何一个数所经过的最右边的位置和最左边的位置差；思路：求每一个数的逆序数，易知最左边的位置一定是出事位置减去该数的逆序数，（自己理解）；最大位置就比较简单了，当前位置和出事位置中最大的哪一个必为该说所能走的最右边位置。

#include <iostream>  #include <string.h>  #include <algorithm>  #include <stdio.h>  #include<math.h>using namespace std;  const int N = 500005;    struct Node  {      int v,order,x;  };    int n;  int c[N];  int aa[N];    //离散化后的数组  Node a[N],t[N];    //树状数组    int Lowbit(int x)  {      return x & (-x);  }    void Update(int t,int val)  {      for(int i=t; i<=n; i+=Lowbit(i))          c[i] += val;  }    int getSum(int x)  {      int ans=0;      for(int i=x; i>0; i-=Lowbit(i))          ans += c[i];      return ans;  }    bool cmp(Node a,Node b)  {      return a.v<b.v;  }    int main()  {      int T,cases=0;    scanf("%d",&T);    while(T--)      {          scanf("%d",&n);        //离散化          for(int i=1; i<=n; i++)          {              scanf("%d",&a[i].v);              a[i].order=i;              t[i]=a[i];        }          sort(a+1,a+1+n,cmp);          for(int i=1; i<=n; i++)              aa[a[i].order]=i;          //树状数组求逆序数          memset(c,0,sizeof(c));          int ans=0;          for(int i=1; i<=n; i++)          {              Update(aa[i],1);              ans=i-getSum(aa[i]);              t[i].x=ans;        }         sort(t+1,t+n+1,cmp);        printf("Case #%d: ",++cases);        for(int i=1;i<n;i++)        {        if(t[i].order>i)          printf("%d ",abs(t[i].x));          else          printf("%d ",abs(i-(t[i].order-t[i].x)));}if(t[n].order>(n))          printf("%d\n",abs(t[n].x));          else          printf("%d\n",abs(n-(t[n].order-t[n].x)));    }      return 0;  }