最令人费解的是题意。

先按照a_i构造一个映射表: map[a_i] = i
再根据b_i取出映射到待排序的数组: c_i = map[b_i]
然后对c数组求逆序对数即可，我的方法是简单的归并排序

代码

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1005;
int n, ans = 0, a[N], c[N];
pair<int, int> b[N];
inline void mergeSort(int l, int r) {
	if (l == r)return;
	int mid = (l + r) >> 1;
	mergeSort(l, mid);
	mergeSort(mid + 1, r);
	int i = l, j = mid + 1, tot = 0;
	while (i <= mid&&j <= r) {
		if (a[i] < a[j])c[++tot] = a[i++];
		else ans += mid - i + 1, c[++tot] = a[j++];
	}
	while (i <= mid)c[++tot] = a[i++];
	while (j <= r)c[++tot] = a[j++];
	for (int t = 0; t < tot; t++)
		a[l + t] = c[1 + t];
}
signed main(void) {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%d%d", &b[i].first, &b[i].second);
	for (int i = 1; i <= n; i++)
		c[b[i].first] = i;
	for (int i = 1; i <= n; i++)
		a[i] = c[b[i].second];
	mergeSort(1, n);
	printf("%d\n", ans);
	system("pause");
}