2016 Multi-University Training Contest 4 1012 hdu 5775(树状数组)

Bubble Sort

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 208    Accepted Submission(s): 139

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.

题解：考虑一个位置上的数字c在冒泡排序过程的变化情况(这个解释我看了很久，也就是说各个数字在冒泡排序的过程中出现的位置，在那些位置编号的最左和最右的绝对值)。c会被其后面比c小的数字各交换一次，之后c就会只向前移动。数组从右向左扫，树状数组维护一下得到每个值右边有多少个比其小的值，加上原位置得到最右位置，最左位置为初始位置和最终位置的最小值。

AC代码：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
#include <string>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <ctime>
#include<complex>
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef vector<int> vi;
#define de(x) cout << #x << "=" << x << endl
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define setIO(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
const int N = 101010;
int T,n,a[N],b[N];
#define lb(x)(x&(-x))
struct Fenwick{
    int a[N],l;
    void add(int x){
        for(int i=x;i<=l;i+=lb(i)) a[i]++;
    }
    int sum(int x){
        int r=0;for(int i=x;i>=1;i-=lb(i)) r+=a[i];return r;
    }
}f;

void main2(){
    scanf("%d",&n);
    rep(i,1,n+1) scanf("%d",a + i);
    f.l = n;fill(f.a,f.a+n+1,0);
    for(int i=n;i>=1;--i) b[a[i]] = i + f.sum(a[i]) - min(a[i] , i) , f.add(a[i]);
    rep(i,1,n+1) printf("%d%c",b[i]," \n"[i==n]);
}

int main(){
  //  freopen("in", "r", stdin);
 //   freopen("out", "w", stdout);
    db t1 = clock();
    scanf("%d",&T);
    rep(i,1,T+1) printf("Case #%d: ",i) , main2();
  //  cerr << (clock() - t1) / CLOCKS_PER_SEC << endl;
    return 0;
}