POJ2299《Ultra-QuickSort》方法：歸併排序找逆序數

題目大意：求一個數的逆序數，比如54321的逆序數爲4+3+2+1+0=10，逆序數 = 在只交換相鄰兩數的前提下，需要的交換次數。

解題思路：因爲這題數據量爲50W，時間限制爲7000ms，用冒泡排序找相鄰交換次數肯定超時，因此用歸併排序找相鄰交換次數。之所以不用快速排序，是因爲它不符合相鄰元素交換的要求。

// 3748k 2454ms #include <iostream> using namespace std; const int inf = 1000000000; __int64 ans; // 逆序數 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數參照，必須要加 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; //要交換大數的交換次數 } } 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 n; while (cin >> n && n!=0) { ans = 0; int *a = new int[n+1]; for (int i = 1;i <= n; ++i) cin >> a[i]; mergeSort(a, 1, n); printf("%I64d\n", ans); //輸出64位的整數，無法用cout delete a; } return 0; }