poj1990 MooFest

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v_i * (x_i * c₁ - x₁) + v_i * (x₂ - x_i * c₂)加入答案中即可。

实现：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MAXN = 200005;
const int MAXX = 200005;
struct node
{
    int v, x;
};
node a[MAXN];
int n;
ll bit0[MAXX], bit1[MAXX];
inline int lowbit(int x) { return x & -x; }
void add(ll * bit, int i, int dx)
{
    while (i <= MAXX) { bit[i] += dx; i += lowbit(i); }
}
ll sum(ll * bit, int i)
{
    ll ans = 0;
    while (i) { ans += bit[i]; i -= lowbit(i); }
    return ans;
}
bool cmp(const node & a, const node & b)
{
    return a.v < b.v;
}
int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d %d", &a[i].v, &a[i].x);
    sort(a, a + n, cmp);
    ll ans = 0;
    for (int i = 0; i < n; i++)
    {
        ll c = sum(bit0, a[i].x), s = sum(bit1, a[i].x);
        ans += (a[i].x * c - s) * a[i].v;
        c = sum(bit0, MAXX) - c; s = sum(bit1, MAXX) - s;
        ans += (s - c * a[i].x) * a[i].v;
        add(bit0, a[i].x, 1); add(bit1, a[i].x, a[i].x);
    }
    printf("%lld
", ans);
    return 0;
}