"BZOJ1132" "POI2008" Tro calculate the area of geometric cross product

Link:

#include <stdio.h>int main(){    puts("转载请注明出处[辗转山河弋流歌 by 空灰冰魂]谢谢");    puts("网址：blog.csdn.net/vmurder/article/details/46605807");}

Exercises

First, violence is O ( n 3 ) Ask for each triangle area!
But how do you ask for triangular area? In general, we use cross-product ... Wait a minute? That one cross product is not to be counted many times?

Well, it is solved, first ordered order to ensure that the order is not heavy, and then calculate the contribution of each cross product, that is, the contribution of each side, and then because of the order of what, the complexity of time O( n 2 Log N) 。

And then this problem. Uh, card accuracy ...?!
To cross the product, the last thing to be divided by 2,, first, not to divide, to the end of a special sentence qwq

Code:

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define N 3010using namespace STD;structpoint{Long Longx, y;voidRead () {scanf("%lld%lld", &x,&y);}} Now,p[n],tp[n];inline Long LongXmul (Point b,point c,point a=now) {return(C.Y-A.Y) * (b.x-a.x)-(B.Y-A.Y) * (c.x-a.x);}BOOLCmpxy (Point A,point B) {returnA.x==b.x? a.y<b.y:a.x<b.x;}BOOLCMPMU (Point A,point B) {returnXmul (b) >0;}intNLong LongAnsintMain () {Freopen ("Test.in","R", stdin);intI,j,k;Long LongSumx,sumy;scanf("%d", &n); for(i=1; i<=n;i++) P[i].read (); Sort (p+1, p+n+1, CMPXY); for(i=1; i<=n-2; i++) {now=p[i]; sumx=sumy=0; for(j=i+1; j<=n;j++) Tp[j]=p[j]; Sort (tp+i+1, tp+n+1, CMPMU); for(j=i+1; j<=n;j++) {sumx+=tp[j].x-now.x;        SUMY+=TP[J].Y-NOW.Y; } for(j=i+1; J <n;j++) {sumx-=tp[j].x-now.x;            SUMY-=TP[J].Y-NOW.Y;            ans+= (tp[j].x-now.x) *sumy;        ans-= (TP[J].Y-NOW.Y) *sumx; }    }if(ans&1)printf("%lld.5\n",ans>>1);Else printf("%lld.0\n",ans>>1);return 0;}

"BZOJ1132" "POI2008" Tro calculate the area of geometric cross product