bzoj-1132 Tro

Test instructions

Given n points, the area of all triangles consisting of these n points is calculated;

n<=3000;

Exercises

This problem O (N^3) enumeration triangle time complexity is unbearable;

So consider enumerating an edge, multiple triangles to calculate, complexity at O (n^2) level;

The triangular area can be used to calculate the height of the area, or the cross-product;

If the method of the bottom multiply height is used to find out the sum of all the points to the straight line distance, it is also possible to get the current solution by O (1);

But the distance and this step must be O (n), it is difficult to transfer to the next bottom up;

And the method of cross product is to simplify the cross product of the formula, you can put the coordinates of points, plus and a calculation;

But the cross product is to have the direction area, this thing card me half a day;

The actual implementation is this:

Enumerates the first point, takes the point set at the top right of the first point, sorts the point set by the slope size of the first point, and then enumerates the second point for solving;

In this way each area is only one time, and because of the slope is orderly, so all triangles are positive area;

Because of the order, Time complexity O (n^2 logn);

(I can't believe I kept two decimal places in WA for one night!!) )

Code:

#include <math.h> #include <stdio.h> #include <string.h> #include <algorithm> #define N 3100using namespace Std;typedef Long long ll;struct point{ll x,y;double slope;friend bool operator < (point A,point b) {i F (a.x==b.x) return A.y<b.y;return a.x<b.x;} Friend point operator-(point A,point B) {return (point) {a.x-b.x,a.y-b.y};} Friend int operator * (Point A,point b) {return a.x*b.y-a.y*b.x;}} A[n],t[n],o;bool CMP (point A,point B) {return a.slope>b.slope;} int main () {int n,m,i,j,k;ll ans,sx,sy,tx,ty;scanf ("%d", &n), for (i=1,sx=sy=0;i<=n;i++) {scanf ("%d%d", &a[i ].X,&AMP;A[I].Y); sx+=a[i].x,sy+=a[i].y;} Sort (a+1,a+1+n); for (i=1,ans=0;i<=n;i++) {sx-=a[i].x,sy-=a[i].y;tx=sx,ty=sy;o=a[i];memcpy (t+i+1,a+i+1,sizeof ( Point) * (N-i)), for (j=i+1;j<=n;j++) {if (t[j].x==o.x) t[j].slope=1e10;elset[j].slope= (double) (T[J].Y-O.Y)/(T[j]. x-o.x);} Sort (t+i+1,t+n+1,cmp); for (j=i+1;j<n;j++) {point v=t[j]-a[i];tx-=t[j].x,ty-=t[j].y;ans+=tx*v.y-ty*v.x+ (n-j) * (A [I].y*v.x-a[i].x*v.y);}} printf"%lld.%d", ans>>1,ans&1?5:0); return 0;} 

bzoj-1132 Tro