POJ1804《Brainman》方法：归并排序找逆序数

题目跟POJ2299差不多，只不过数据集变小了，用冒泡排序也可以。

// 204k 204ms#include <iostream>using namespace std;const int inf = 1000001;int ans; // 逆序数void combine(int *a, int top, int mid, int end){int len1 = mid-top+1;int len2 = end-mid;int *left = new int[len1+2];int *right = new int[len2+2];int i, j, k;for (i = 1; i <= len1; ++i)left[i] = a[top-1+i];left[len1+1] = inf; //inf数参照，必须要加 for (j = 1; j <= len2; ++j)right[j] = a[mid+j];right[len2+1] = inf;i = j = 1;for (k=top; k <= end; ) {if (left[i] <= right[j]) {a[k++] = left[i++];} else {a[k++] = right[j++];ans += len1-i+1; //要交换大数的交换次数}}delete left;delete right;}void mergeSort(int *a, int top, int end) {if (top < end) {int mid = (top+end) / 2;mergeSort(a, top, mid);mergeSort(a, mid+1, end);combine(a, top, mid, end);}}int main(){freopen("temp.txt", "r", stdin);int cases, n, num = 0;cin >> cases;while (cases--) {cin >> n;ans = 0;int *a = new int[n+1];for (int i = 1;i <= n; ++i) cin >> a[i];mergeSort(a, 1, n);cout << "Scenario #" << ++num << ":" <<endl;cout << ans << endl << endl;delete a;}return 0;}