HDU 1394（归并求逆序对）

HDU 1394

原始序列逆序对为ans，新序列的逆序对减少s[i]个，增加n - (s[i] + 1)个，所以增加的逆序对为ans += n - 1 - 2*s[i]；

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N  = 5e5+10;
int s[N], b[N], tmp[N];
ll cnt;
inline  void merge_sort(int a[], int l, int r){if(l == r) return;int mid = (l + r) / 2;merge_sort(a, l, mid);merge_sort(a, mid + 1, r);int i = l, j = mid + 1, k = 0;while(i <= mid && j <= r){if(a[i] <= a[j]) tmp[k++] = a[i++];else tmp[k++] = a[j++], cnt += (ll)mid - i + 1;}while(i <= mid) tmp[k++] = a[i++];while(j <= r) tmp[k++] = a[j++];for(i = 0; i < k; i++){a[l + i] = tmp[i];}
}
int main(){
#ifdef ONLINE_JUDGE
#elsefreopen("in.txt", "r", stdin);
#endif //ONLINE_JUDGEint n;while(~scanf("%d", &n)){cnt = 0;for(int i = 1; i <= n; i++){scanf("%d", &s[i]);b[i] = s[i];}merge_sort(b, 1, n);ll ans = 0x3f3f3f3f;for(int i = 1; i <= n; i++){cnt += (n - 1 - 2 * s[i]);ans = min(cnt, ans);}printf("%lld\n", ans);}return 0;
}