HOJ2098//归并排序求逆系数

Ultra-QuickSort

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 Source : Waterloo ACM Programming Contest Feb 5, 2005 Time limit : 3 sec Memory limit : 32 M

Submitted : 519, Accepted : 171

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

59105431230

Sample Output

60

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#include<iostream>#include<stdio.h>#define MAX  500000using namespace std;int a[MAX +1],t[MAX +1];long long nxs;void Merge(int left, int mid, int right)//这个是照数据结构上面的模版写的，感觉写的很清晰,以后就用它做模版啦{    int i = left, j = mid + 1,k=0;    while(i <= mid && j <= right)    {        if (a[i] > a[j])        {            t[k++] = a[j++];            nxs += mid - i +1;// a[i]后面的数字对于a[j]都是逆序的        }        else        {            t[k++] = a[i++];        }    }    while(i <= mid) t[k++] = a[i++];    while(j <= right) t[k++] = a[j++];    for (i = 0; i < k; i++)//这一步是必须的，归并完成后将结果复制到原输入数组    {      a[left+i]=t[i];    }}void MergeSort(int left, int right){    int mid;    if (left < right)    {        mid = (left + right) / 2;        MergeSort(left, mid);        MergeSort(mid + 1, right);        Merge(left, mid, right);    }}int main(){    int i,n;    while(scanf("%d", &n)&&n)    {        nxs=0;        for(i = 0; i < n; i++)        {            scanf("%d", &a[i]);        }        MergeSort(0, n - 1);        printf("%lld\n", nxs);    }    return 0;}

主要就是利用归并排序求逆系数，以前没有学过排序，思路都是别人的，当作学习吧，嘻嘻！