A.归并排序

归并排序 （求逆序数）

归并排序：递归+合并+排序

时间复杂度：O（n logn）    空间复杂度：O（n）

用途：1.排序  2.求逆序对数

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 sequence9 1 0 5 4 ,        Ultra-QuickSort produces the output0 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：
给出长度为n的无序序列，每次只能交换相邻的两个数，求至少要交换多少次才使得该序列为递增序列。

分析：
1.根据时间范围估计时间复杂度，归并排序的时间复杂度一般为O（n logn）
2.算法步骤： 1.划分问题：把序列分为元素个数尽量相等的两半
           2.递归求解：把两半元素分别排序
           3.合并问题：把两个有序表合成一个
3.保存时，用long long 类型，否则会wa的

代码：

#include<cstdio>
#include<iostream>
using namespace std;
const int maxn=500001;

int a[maxn],b[maxn];
long long ans;

void merge(int l,int m,int r)       //归并排序
{
    int i=l,j=m+1,t=0;
    while(i<=m&&j<=r)   //从小到大排序
    {
        if(a[i]>a[j])
        {
            b[t++]=a[j++];
            ans+=m-i+1;
        }
        else b[t++]=a[i++];
    }
    while(i<=m)
        b[t++]=a[i++];
    while(j<=r)
        b[t++]=a[j++];
    for(int i=0;i<t;i++)   //合并
        a[l+i]=b[i];
}

void merge_sort(int l,int r)  //划分问题
{
    if(l<r)
    {
        int m=(l+r)/2;
        merge_sort(l,m);
        merge_sort(m+1,r);
        merge(l,m,r);
    }
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        ans=0;
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        merge_sort(0,n-1);
        printf("%lld\n",ans);
    }
    return 0;
}

本来想着完全不看别人的代码都由自己做，可是看了一天了只是有了思路，还是写不出来。。。又看来了别人的代码。。。