思路：
首先把牛按v值排序，然后对于每一头牛i，分别统计v值小于该牛且（1）x值小于该牛的牛数c1以及这些牛的坐标和x1；（2）x值大于该牛的牛数c2以及这些牛的坐标和x2。将vi * (xi * c1 - x1) + vi * (x2 - xi * c2)加入答案中即可。
实现：
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 using namespace std; 5 typedef long long ll; 6 const int MAXN = 200005; 7 const int MAXX = 200005; 8 struct node 9 { 10 int v, x; 11 }; 12 node a[MAXN]; 13 int n; 14 ll bit0[MAXX], bit1[MAXX]; 15 inline int lowbit(int x) { return x & -x; } 16 void add(ll * bit, int i, int dx) 17 { 18 while (i <= MAXX) { bit[i] += dx; i += lowbit(i); } 19 } 20 ll sum(ll * bit, int i) 21 { 22 ll ans = 0; 23 while (i) { ans += bit[i]; i -= lowbit(i); } 24 return ans; 25 } 26 bool cmp(const node & a, const node & b) 27 { 28 return a.v < b.v; 29 } 30 int main() 31 { 32 scanf("%d", &n); 33 for (int i = 0; i < n; i++) scanf("%d %d", &a[i].v, &a[i].x); 34 sort(a, a + n, cmp); 35 ll ans = 0; 36 for (int i = 0; i < n; i++) 37 { 38 ll c = sum(bit0, a[i].x), s = sum(bit1, a[i].x); 39 ans += (a[i].x * c - s) * a[i].v; 40 c = sum(bit0, MAXX) - c; s = sum(bit1, MAXX) - s; 41 ans += (s - c * a[i].x) * a[i].v; 42 add(bit0, a[i].x, 1); add(bit1, a[i].x, a[i].x); 43 } 44 printf("%lld\n", ans); 45 return 0; 46 }
