python/TangDou/AcWing/BIT/P2345.cpp

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 20010;

// 听力值v,坐标x

// 结构体+第一维、第二维由小到大排序
struct Node {
    int v, x;
    const bool operator<(const Node &t) const {
        if (v == t.v) return x < t.x;
        return v < t.v;
    }
} a[N];

// 树状数组模板
int c1[N], c2[N];

#define lowbit(x) (x & -x)

void add(int c[], int x, int d) {
    for (int i = x; i < N; i += lowbit(i)) c[i] += d;
}

LL sum(int c[], int x) {
    LL res = 0;
    for (int i = x; i; i -= lowbit(i)) res += c[i];
    return res;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("P2345.in", "r", stdin);
#endif

    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d %d", &a[i].v, &a[i].x);

    sort(a + 1, a + n + 1); // 排序

    LL res = 0;
    for (int i = 1; i <= n; i++) {
        // c1:坐标和树状数组
        // c2:牛的个数
        LL s1 = sum(c1, a[i].x - 1);              // a[i]进入树状数组时，它前面所有牛的坐标和
        LL s2 = sum(c1, 20000) - sum(c1, a[i].x); // a[i]进入树状数组时，它后面所有牛的坐标和

        LL cnt = sum(c2, a[i].x);
        /*
        cnt:它之前牛的个数
        i-1-cnt:它之后牛的个数
        a[i].x：它的坐标
        */

        // 方法1：cout << "i=" << i << ",sum(c2,N)=" << sum(c2, N) << endl;
        res += a[i].v * (cnt * a[i].x - s1 + s2 - (sum(c2, N) - cnt) * a[i].x);

        //  方法2：
        // res += a[i].v * (cnt * a[i].x - s1 + s2 - (i - 1 - cnt) * a[i].x);

        add(c1, a[i].x, a[i].x); // 维护坐标前缀和
        add(c2, a[i].x, 1);      // 维护个数前缀和
    }
    // 输出结果
    printf("%lld\n", res);
    return 0;
}