题目链接：https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1107
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<bits/stdc++.h>
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;
}
原文：http://www.cnblogs.com/westwind1005/p/6352990.html