hdu5755 Bubble Sort（树状数组or线段树）

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本文介绍了一种使用树状数组优化冒泡排序的方法，通过跟踪数组中每个元素的位置变化来确定其在排序过程中所能达到的最左与最右位置，进而计算每个数值在其路径上左右边界间的绝对距离。

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思路：考虑一个位置上的数字c在冒泡排序过程的变化情况。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。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>

#define INF 0x3f3f3f3f
#define esp 1e-9
typedef long long LL;
using namespace std;
const int maxn = 100000 +100;
int c[maxn], a[maxn], rid[maxn], ans[maxn];
struct Node
{
    int l, r, v;
}st[maxn*8];
void Build(int x, int L, int R)
{
    st[x].l=L;
    st[x].r=R;
    st[x].v=0;
    if(L+1==R) return;
    int m = (L+R)/2;
    Build(x<<1, L, m);
    Build(x<<1|1, m, R);
}
void push_down(int x)
{
    st[x<<1].v += st[x].v;
    st[x<<1|1].v += st[x].v;
    st[x].v = 0;
}
void Update(int x, int L, int R, int v)
{
    if(L<=st[x].l&&st[x].r<=R)
    {
        st[x].v+=v;
        return;
    }
    push_down(x);
    int m = (st[x].l+st[x].r)/2;
    if(L<m) Update(x<<1, L, R, v);
    if(m<R) Update(x<<1|1, L, R, v);
}
int Queue(int x, int L, int R)
{
    if(L<=st[x].l&&st[x].r<=R)
    {
        return st[x].v;
    }
    push_down(x);
    int ans = 0;
    int m = (st[x].l +st[x].r)/2;
    if(L<m) ans += Queue(x<<1, L, R);
    if(m<R) ans += Queue(x<<1|1, L, R);
    return ans;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T, cas=1, n;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        for(int i=1;i<=n;++i)
        {
            scanf("%d", &a[i]);
            rid[a[i]]=i;
        }
        Build(1, 1, n+1);
        for(int i=1;i<=n;++i)
        {
            int l = min(i, rid[i]);
            int r = Queue(1, rid[i], rid[i]+1)+rid[i];
    //        cout<<"l:"<<l<<" r:"<<r<<endl;
            Update(1, 1, rid[i]+1, 1);
            ans[i] = abs(l-r);
        }
        printf("Case #%d: ", cas++);
        for(int i=1;i<=n;++i)
        {
            printf("%d%c", ans[i], i==n?'\n':' ');
        }
    }
    return 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

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