poj-2299-Ultra-QuickSort(线段树 || 归并排序)

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：求n个数经过冒泡排序变成有序后，交换的次数

线段树
记录每个数和它开始的位置pos，然后将它们排序，最后计算pos中有多少逆序对（类似于poj-3067-Japan）

#include <iostream> 
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm> 
#define N 500005
#define ll long long
using namespace std;
int n;
ll ans[N];
struct node
{
    int pos, num;
    friend bool operator < (node a, node b)
    {
        return a.num < b.num;
    }
}aaa[N];
void update(int x, int val)
{
    while(x <= n)
    {
        ans[x] += val;
        x += x&-x;
    }
}
ll sum(int x)
{
    ll ret = 0;
    while (x > 0)
    {
        ret += ans[x];
        x -= x&-x;
    }
    return ret;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    int i, j, K, T;
    long long ret;
    while(scanf("%d", &n), n)
    {
        ret = 0;
        memset(ans, 0, sizeof(ans));
        for (i = 0; i < n; i++)
        {
            scanf("%d", &aaa[i].num);
            aaa[i].pos = 1+i;
        }
        sort(aaa, aaa+n);
        for (i = 0; i < n; i++)
        {
            ret += sum(n)-sum(aaa[i].pos);
            update(aaa[i].pos, 1);
        }
        printf("%lld\n", ret);
    }
    return 0;
}

归并排序

#include <iostream> 
#include <cstdio>
#include <cstdlib>
#include <cstring>
#define N 500010
using namespace std;
int n, a[N], t[N];
long long ans;
void Merg_Sort(int l, int r)
{
    if (l >= r-1)   return ;
    int m = (l+r)>>1;
    Merg_Sort(l, m);
    Merg_Sort(m, r);
    int p = l, q = m, i = l;
    while(p < m || q < r)
    {
        if (q >= r || (p < m && a[p] < a[q]))   t[i++] = a[p++];
        else
        {
            if (p < m)  ans += (m-p);
            t[i++] = a[q++];
        }
    }
    for (i = l; i < r; i++)
        a[i] = t[i];
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    int i;
    while(scanf("%d", &n), n)
    {
        memset(a, 0, sizeof(a));
        memset(t, 0, sizeof(t));
        for (i = 0; i < n; i++)
            scanf("%d", &a[i]);
        ans = 0;
        Merg_Sort(0, n);
        printf("%lld\n", ans);
    }
    return 0;
}