70 lines
1.8 KiB
C++
70 lines
1.8 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int N = 20010;
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// 听力值v,坐标x
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// 结构体+第一维、第二维由小到大排序
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struct Node {
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int v, x;
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const bool operator<(const Node &t) const {
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if (v == t.v) return x < t.x;
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return v < t.v;
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}
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} a[N];
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// 树状数组模板
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int c1[N], c2[N];
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#define lowbit(x) (x & -x)
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void add(int c[], int x, int d) {
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for (int i = x; i < N; i += lowbit(i)) c[i] += d;
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}
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LL sum(int c[], int x) {
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LL res = 0;
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for (int i = x; i; i -= lowbit(i)) res += c[i];
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return res;
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("P2345.in", "r", stdin);
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#endif
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int n;
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scanf("%d", &n);
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for (int i = 1; i <= n; i++) scanf("%d %d", &a[i].v, &a[i].x);
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sort(a + 1, a + n + 1); // 排序
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LL res = 0;
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for (int i = 1; i <= n; i++) {
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// c1:坐标和树状数组
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// c2:牛的个数
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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
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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
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LL cnt = sum(c2, a[i].x);
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/*
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cnt:它之前牛的个数
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i-1-cnt:它之后牛的个数
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a[i].x：它的坐标
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*/
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// 方法1：cout << "i=" << i << ",sum(c2,N)=" << sum(c2, N) << endl;
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res += a[i].v * (cnt * a[i].x - s1 + s2 - (sum(c2, N) - cnt) * a[i].x);
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// 方法2：
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// res += a[i].v * (cnt * a[i].x - s1 + s2 - (i - 1 - cnt) * a[i].x);
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add(c1, a[i].x, a[i].x); // 维护坐标前缀和
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add(c2, a[i].x, 1); // 维护个数前缀和
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}
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// 输出结果
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printf("%lld\n", res);
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return 0;
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}
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