poj 2299 Ultra-QuickSort

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

Source

Waterloo local 2005.02.05

 

 

算法课学过归并排序求逆序数。

代码：

#include <iostream>
#include <cstdio>
#include <set>

using namespace std;
typedef long long ll;
int num[500000];
ll merge_sort(int *t,int l,int r) {
    if(l >= r) return 0;
    int mid = (l + r) / 2;
    ll d = merge_sort(t,l,mid) + merge_sort(t,mid + 1,r);
    int *temp = new int[r - l + 1];
    int c = 0,i = l,j = mid + 1;
    while(i <= mid || j <= r) {
        if(i <= mid && (t[i] <= t[j] || j > r)) {
            temp[c ++] = t[i ++];
        }
        else if(j <= r && (t[j] < t[i] || i > mid)) {
            temp[c ++] = t[j ++];
            d += mid - i + 1;
        }
    }
    for(int i = l;i <= r;i ++) {
        t[i] = temp[i - l];
    }
    delete [] temp;
    return d;
}
int main() {
    int n,d;
    while(~scanf("%d",&n) && n) {
        for(int i = 0;i < n;i ++) {
            scanf("%d",&num[i]);
        }
        printf("%lld
",merge_sort(num,0,n - 1));
    }
}

耗时不少，其实树状数组好一点，只不过不能直接把数作为下标，数很大，但是个数最多500000，可以接受，可以排序后进行。

代码：

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

using namespace std;
typedef long long ll;
typedef pair<int,int> pa;
pa num[500000];
ll sum[500001];
int n;
inline int lowbit(int x) {
    return x & -x;
}
inline void update(int x,int y) {
    while(x <= n) {
        sum[x] += y;
        x += lowbit(x);
    }
}
inline int getsum(int x) {
    int ans = 0;
    while(x > 0) {
        ans += sum[x];
        x -= lowbit(x);
    }
    return ans;
}
bool cmp(pa a,pa b) {
    if(a.first == b.first) return a.second < b.second;
    return a.first < b.first;
}
int main() {
    int d;
    while(~scanf("%d",&n) && n) {
        ll ans = 0;
        for(int i = 0;i < n;i ++) {
            sum[i + 1] = 0;
            scanf("%d",&d);
            num[i] = pa(d,i + 1);
        }
        sort(num,num + n,cmp);
        for(int i = 0;i < n;i ++) {
            ans += getsum(num[i].second);
            update(num[i].second,1);
        }
        printf("%lld
",(ll)n * (n - 1) / 2 - ans);
    }
}