CSU-ACM2017暑期訓練16-樹狀數組 A - Ultra-QuickSort POJ - 2299

A - Ultra-QuickSort

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence
9 1 0 5 4 ,

produces the output
0 1 4 5 9 .

Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence. 

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed. 

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence. 

Sample Input

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0

利用樹狀數組計算逆序數即可，注意數據類型的取值範圍。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
typedef long long ll;
using namespace std;
const ll maxn = 5e5+19;

int n;
ll A[maxn], B[maxn], C[maxn];

void add(ll x, ll d){
    while(x <= n){
        C[x] += d; x += x&-x;
    }
}

ll sum(ll x){
    ll ret = 0;
    while(x > 0){
        ret += C[x]; x -= x&-x;
    }
    return ret;
}

int main(){
#ifdef TEST
freopen("test.txt", "r", stdin);
#endif // TEST

    while(cin >> n){
        if(n == 0)
            break;
        memset(A, 0, sizeof(A));
        memset(B, 0, sizeof(B));
        memset(C, 0, sizeof(C));
        for(int i = 1; i <= n; i++){
            scanf("%lld", &A[i]);
            B[i] = A[i];
        }
        // 對A[]離散化。
        sort(B+1, B+n+1);
        unique(B+1, B+n+1);
        for(int i = 1; i <= n; i++){
            A[i] = lower_bound(B+1, B+n+1, A[i]) - B;
        }
        ll res = 0;
        for(int i = 1; i <= n; i++){
            add(A[i], 1);
            res += i-sum(A[i]);
        }
        printf("%lld\n", res);
    }

    return 0;
}