Ultra-QuickSort（树状数组 + 离散化）

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 39782		Accepted: 14340

Description

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 ,
Ultra-QuickSort 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

 

      题意：

      给出数 n （0 ~ 500000）代表有 n 个数，后给出这 n 个数（0 ~ 999999999），求出需要交换的次数使这个序列变为递增序列。

 

      思路：

      树状数组 + 离散化。离散化用 排序 + map 会导致 TLE，所以要用 排序 + 去重 + 二分 来离散化，结果应该要为 long long 型的。题目即为求逆序数，运用树状数组即可求出。

 

      AC：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

const ll MAX = 500005;

ll num[MAX], s_num[MAX];
ll bit[MAX], n;

void add(ll i, ll x) {

        while (i <= n) {
                bit[i] += x;
                i += i & -i;
        }

}

ll sum(ll i) {
        ll s = 0;

        while (i > 0) {
                s += bit[i];
                i -= i & -i;
        }

        return s;
}

int main() {

        while (~scanf("%I64d", &n) && n) {

                for (ll i = 0; i < n; ++i) {
                        scanf("%I64d", &num[i]);
                        s_num[i] = num[i];
                        bit[i + 1] = 0;
                }

                sort(s_num, s_num + n);

                ll ans = 0;
                for (ll i = 0; i < n; ++i) {
                        int in = lower_bound(s_num, s_num + n, num[i]) - s_num;

                        ans += i - sum(in);
                        add(in + 1, 1);
                }

                printf("%I64d\n", ans);

        }


        return 0;
}