题目链接：

思路：其实就是升级版的逆序数，x坐标当作位置，y坐标当作数值val，只是可能有相等的数，稍作修改即可。

#include
     
      using namespace std;typedef long long ll;const int N = 5e4 + 3;int lowbit(int x){    return x & (-x);}struct node{    int x,y,order;}q[N];int a[N],c[N],n;bool cmp1(node t1,node t2){    if(t1.x != t2.x)        return t1.x < t2.x;    return t1.y < t2.y;}bool cmp2(node t1,node t2){    if(t1.y != t2.y)        return t1.y < t2.y;    return t1.order < t2.order;}void update(int pos,int val){    for(int i = pos; i <= n; i += lowbit(i))        c[i] += val;}int sum(int pos){    int ans = 0;    for(int i = pos; i > 0; i -= lowbit(i))        ans += c[i];    return ans;}int main(){    scanf("%d",&n);    for(int i = 1; i <= n; i++)        scanf("%d %d",&q[i].x,&q[i].y);    sort(q+1,q+n+1,cmp1);    for(int i = 1; i <= n; i++)        q[i].order = i;    sort(q+1,q+n+1,cmp2);    for(int i = 1; i <= n; i++)        a[q[i].order] = i;    int ans = 0;    for(int i = 1; i <= n; i++)    {        update(a[i],1);        ans += i - sum(a[i]);    }    printf("%d\n",ans);    return 0;}