标签:int() out put oid cut ios point int inpu
比赛的时候想不出思路 很难受
#include<iostream> #include <stdio.h> #include <limits.h> #define LL long long using namespace std; const int maxn= (int)1e5+10; struct point{ LL x,y; point(){} point(LL x,LL y):x(x),y(y){} void input(){scanf("%I64d%I64d",&x,&y);} void output(){cout<<x<<" "<<y<<endl;} point operator - (const point& a){return point(x-a.x,y-a.y);} LL operator * (const point &a){return x*a.y-y*a.x;} }; point p[maxn]; template <class T> T abs(T a){return a>=0? a:-a;} LL getvol(int n){ LL res=0; for(int i=1;i+1<n;i++) res+=(p[i+1]-p[0])*(p[i]-p[0]); return abs(res); } int n; int main() { #ifdef shuaishuai freopen("a.txt","r",stdin); #endif // shuaishuai scanf("%d",&n); for(int i=0;i<n;i++){ p[i].input(); } LL ans=-1; LL sum=getvol(n); LL now=abs( (p[2]-p[0])*(p[1]-p[0]) ); int j=2; for(int i=0;i<n;i++){ while(1){ LL tmp=now+abs((p[(j+1)%n]-p[i])*(p[j]-p[i])); if(abs(sum-tmp-tmp)>abs(sum-now-now)) break; //printf("%I64d %I64d %I64d i=%d j=%d\n",sum,tmp,now,i,j); j=(j+1)%n; now=tmp; } ans=max(ans,max(sum-now,now)); // cout<<ans<<endl; now-=abs((p[j]-p[i])*(p[(i+1)%n]-p[i])); } printf("%I64d %I64d\n",ans,sum-ans); return 0; }
标签:int() out put oid cut ios point int inpu
原文地址:http://www.cnblogs.com/MeowMeowMeow/p/7291256.html