Ultra-QuickSort （歸併排序，逆序數）

標籤：逆序對   several   c++   sed   freopen   ace   ==   mina   cout   

題目描述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.

 

輸入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.

 

輸出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.

 

範例輸入

59105431230

 

範例輸出

60

#include <bits/stdc++.h>using namespace std;const int maxn = 5e5 + 1000;const int inf = 0x3f3f3f3f;int s[maxn], t[maxn];long long cnt = 0;int n;void merge(int l1, int r1, int l2, int r2) {    int k = 0;    while (l1 <= r1 && l2 <= r2) {        if (s[l2] < s[l1]) t[++k] = s[l2++], cnt += r1 - l1 + 1;            //若s[l2]小，則l1~r1都比a[j]要大，他們都會與a[j]構成逆序對        else t[++k] = s[l1++];    }    while (l1 <= r1) t[++k] = s[l1++];    while (l2 <= r2) t[++k] = s[l2++];    for (int i = k, j = r2; i >= 1; i--, j--) {//j的最終值不一定為1~r2        s[j] = t[i];    }}void mergesort(int l, int r) {    int mid = (l + r) / 2;    if (l < r) {        mergesort(l, mid);        mergesort(mid + 1, r);        merge(l, mid, mid + 1, r);    }}int main() {    freopen("input.txt", "r", stdin);    while (cin >> n) {        cnt = 0;        if (n == 0)            break;        for (int i = 1; i <= n; i++)            cin >> s[i];        mergesort(1, n);        cout << cnt << endl;    }    return 0;}

 

 

Ultra-QuickSort （歸併排序，逆序數）