【HDU - 5775】【Bubble Sort】

题目：

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. iutput 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.   Author FZU   Source 2016 Multi-University Training Contest 4

 解题思路：树状数组求解，找每位数的右侧又多少比它小的，就是这个数可以到达的最右的位置。

ac代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define maxn 1000000
using namespace std;
int n;
struct node{
	int num;
	int pos;
}nn[maxn];
bool cmp(node a,node b)
{
	return a.num<b.num;
}
int l[maxn],r[maxn];
int c[maxn],pre[maxn];
int lowbit(int x)
{
	return x&-x;
}
void updata(int x,int val)
{
	while(x<=n)
	{
		c[x]+=val;
		x+=lowbit(x);
	}
} 
int query(int x)
{
	int ans=0;
	while(x>0)
	{
		ans+=c[x];
		x-=lowbit(x);
	}
	return ans;
}

int main()
{
	int t;
	cin>>t;
	for(int kase=1;kase<=t;kase++)
	{
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&nn[i].num);	
			pre[nn[i].num]=i;
			nn[i].pos=i;
			l[nn[i].num]=r[nn[i].num]=i;
		}
		sort(nn+1,nn+n+1,cmp);
		memset(c,0,sizeof(c));
		for(int i=1;i<=n;i++)
		{
			int tmp=i-1-query(nn[i].pos);
			r[i]=max(pre[i]+tmp,r[i]);
			l[i]=min(l[i],i);
			updata(nn[i].pos,1);
		}
		printf("Case #%d: ",kase);
		for(int i=1;i<n;i++)
			printf("%d ",abs(r[i]-l[i]));
		printf("%d\n",abs(r[n]-l[n]));	
	}
	return 0;
}