UVALive 7068 K.Bro Sorting（树状数组求逆序对）

UVALive - 7068 K.Bro Sorting

Description

Matt’s friend K.Bro is an ACMer.
Yesterday, K.Bro learnt an algorithm: Bubble sort. Bubble sort will compare each pair of adjacent items and swap them if they are in the wrong order. The process repeats until no swap is needed.
Today, K.Bro comes up with a new algorithm and names it K.Bro Sorting.
There are many rounds in K.Bro Sorting. For each round, K.Bro chooses a number, and keeps swapping it with its next number while the next number is less than it.
For example, if the sequence is “1 4 3 2 5”, and K.Bro chooses “4”, he will get “1 3 2 4 5” after this round. K.Bro Sorting is similar to Bubble sort, but it’s a randomized algorithm because K.Bro will choose a random number at the beginning of each round. K.Bro wants to know that, for a given sequence, how many rounds are needed to sort this sequence in the best situation. In other words, you should answer the minimal number of rounds needed to sort the sequence into ascending order. To simplify the problem, K.Bro promises that the sequence is a permutation of 1, 2, . . . , N.

Input

The first line contains only one integer T (T ≤ 200), which indicates the number of test cases. For each test case, the first line contains an integer N (1 ≤ N ≤ 106).
The second line contains N integers ai (1 ≤ ai ≤ N), denoting the sequence K.Bro gives you. The sum of N in all test cases would not exceed 3 × 106.

Output

For each test case, output a single line ‘Case #x: y’, where x is the case number (starting from 1), y is the minimal number of rounds needed to sort the sequence.

Hint

In the second sample, we choose “5” so that after the first round, sequence becomes “1 2 3 4 5”, and the algorithm completes.

Sample Input

2
5
5 4 3 2 1
5
5 1 2 3 4

Sample Output

Case #1: 4
Case #2: 1

题解

题意

对一个序列进行冒泡排序，求所需要的交换次数

思路

冒泡排序之中之所以需要交换，原因在于存在逆序对，即每当存在逆序对时，便需要进行一次交换。

所以利用树状数组进行维护，按照序列顺序每次add，然后getsum之前的值即可。

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;

const int MAXN = 1e6+10;
int C[MAXN],a[MAXN];
int N; 

int lowbit(int x){
	return x&(-x);
}

void update(int x,int val){
	for(int i =x;i<MAXN;i+=lowbit(i)){
		C[i]+=val;
	}
}

int getsum(int x){
	int ans = 0;
	for(int i=x;i>=1;i-=lowbit(i)){
		ans+=C[i];
	}
	return ans;
}

void init(){
	memset(C,0,sizeof(C));
}

int main() {
    int T;
    while(~scanf("%d",&T)){
		int kase = 0;
		while(T--){
			init();
			kase++;
			scanf("%d",&N);
			int ans = 0;
			for(int i=0;i<N;i++)	scanf("%d",&a[i]);
			for(int i=0;i<N;i++)	{
				update(a[i],1);
				if(getsum(a[i])!=a[i])	ans++;
			}
			printf("Case #%d: %d
",kase,ans);
		}
	}
    return 0;
}