POJ-2299（归并逆序）

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 51525		Accepted: 18911

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：给一串序列，求逆序数
思路：归并

#include <iostream>
#include <cstdio>
using namespace std;

#define maxn 500010
int n, num;

//要归并的序列
int seq[maxn];
//暂存序列
int str[maxn];
//逆序数
long long count;

//归并排序的子排序，将2个穿归并成一个，求逆序数
template<typename T>
long long mergeCountNum(T str1, int size1, T str2, int size2)
{
    int i,j,k;
    long long num;
    i = j = k = num = 0;

    //取2个串中较小元，组成总串
    while(i < size1 && j < size2){
        if(str1[i] <= str2[j]){
            str[k++] = str1[i++];
            //加上逆序数
            num += j;
        }
        else str[k++] = str2[j++];
    }

    //将剩余的元素合并到总串后
    while (i < size1){
        str[k++] = str1[i++];
        //加上逆序数
        num += j;
    }
    while (j < size2)
        str[k++] = str2[j++];

    //拷贝总串
    for (int x = 0; x < (size1 + size2); x++)
        str1[x] = str[x];

    return num;
}

//分治归并排序并返回逆序数
template<typename T>
long long mergesort(T str, int size)
{
    if(size <= 1) return 0;
    T str1 = str;
    int size1 = size / 2;
    T str2 = str + size1;
    int size2 = size - size1;
    return mergesort(str1,size1) + mergesort(str2,size2)
        + mergeCountNum(str1,size1,str2,size2);
}

int main()
{
    while(scanf("%d", &n), n){
        for(int i = 0; i < n; i++){
            scanf("%d", seq+i);
        }
        count = mergesort(seq, n);
        printf("%lld\n", count);
    }
    return 0;
}