思路
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Runtime Error：读入时+10000，防止work数组内坐标为负数
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暴力想到枚举每个三角形直角顶点，优化发现：（截图来自洛谷题解）
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想到前缀和三角形面积，比如第一个sort+work，处理以上图x节点为直角顶点的三角形，ans每次加的为以x，y为顶点的前缀面积
代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define mod 1000000007
#define maxn 100005
using namespace std;
int n,cnt_x[maxn],cnt_y[maxn],last_x[maxn],last_y[maxn];
long long ans,sum_x[maxn],sum_y[maxn];
struct fdfdfd{int x,y;}a[maxn];
bool cmp1(fdfdfd a,fdfdfd b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}
bool cmp2(fdfdfd a,fdfdfd b){return a.x<b.x||(a.x==b.x&&a.y>b.y);}
bool cmp3(fdfdfd a,fdfdfd b){return a.x>b.x||(a.x==b.x&&a.y<b.y);}
bool cmp4(fdfdfd a,fdfdfd b){return a.x>b.x||(a.x==b.x&&a.y>b.y);}
void work()
{
memset(sum_x,0,sizeof(sum_x)); memset(sum_y,0,sizeof(sum_y));
memset(cnt_x,0,sizeof(cnt_x)); memset(cnt_y,0,sizeof(cnt_y));
for(int i=1;i<=n;++i)
{
int x=a[i].x,y=a[i].y;
sum_x[x]=(sum_x[x]+(long long)abs(y-last_x[x])*cnt_x[x])%mod;
++cnt_x[x]; last_x[x]=y;
sum_y[y]=(sum_y[y]+(long long)abs(x-last_y[y])*cnt_y[y])%mod;
++cnt_y[y]; last_y[y]=x;
ans=(ans+sum_x[x]*sum_y[y])%mod;
}
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;++i) scanf("%d%d",&a[i].x,&a[i].y),a[i].x+=10000,a[i].y+=10000;
sort(a+1,a+n+1,cmp1); work();
sort(a+1,a+n+1,cmp2); work();
sort(a+1,a+n+1,cmp3); work();
sort(a+1,a+n+1,cmp4); work();
printf("%lld
",ans);
return 0;
}