HDU5775 Bubble Sort树状数组

题目链接：HDU5775

Bubble Sort

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 653    Accepted Submission(s): 387

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.
 

 

题意：对于1到n的一种序列，使用题干的冒泡排序法排序，问对于每个数排序过程中到达的最右端与最左端相差多少。

题目分析：不难发现，对于一个数i，其运动轨迹是先向右移再向左移，原数组i右面有几个比他小的就右移多少格，最左端由i原来的位置和最后的位置最小值决定。

所以最左端O（1）统计，最右端用树状数组优化，时间复杂度O（nlogn）。

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int s[100010],p[100010];int r[100011];int c[100010];int n,t;int lowbit(int x){    return x&(-x);}void update(int x,int num){    while(x<=n)    {        c[x]+=num;        x+=lowbit(x);    }}int getsum(int x){    int s=0;    while(x>0)    {        s+=c[x];        x-=lowbit(x);    }    return s;}int main(){    cin>>t;    for(int k=1;k<=t;k++)    {        scanf("%d",&n);        memset(r,0,sizeof(r));        memset(c,0,sizeof(c));        for(int i=1;i<=n;i++)        {            scanf("%d",&s[i]);            p[s[i]]=i;        }        for(int i=n;i>=1;i--)        {            r[s[i]]=getsum(s[i]);            update(s[i],1);        }        printf("Case #%d:",k);        for(int i=1;i<=n;i++)        {            printf("% d",p[i]+r[i]-min(p[i],i));        }        printf("\n");    }}