ACM-ICPC国际大学生程序设计竞赛北京赛区(2017)网络赛 题解汇总 Territorial Dispute
A:模拟水题
http://blog.csdn.net/axuhongbo/article/details/78074877
E：计算几何，比赛时候队友暴力枚举过去的，但觉得凸包做更普遍一些，贴个凸包模板
http://www.cnblogs.com/Just–Do–It/p/7582228.html
这个是比较清楚一点的
http://blog.csdn.net/creatorx/article/details/78072775
下面是我队友写的
#include return 0;
if(x<0)return -1;
return 1;
}
struct point
{
double x,y;
};
struct line
{
point s,e;
};
double mul(point sp,point ep,point op)
{
return ((sp.x-op.x)*(ep.y-op.y)-(ep.x-op.x)*(sp.y-op.y));
}
bool inter(line u ,line v)
{
return ((max(u.s.x,u.e.x)>=min(v.s.x,v.e.x))&&
(max(v.s.x,v.e.x)>=min(u.s.x,u.e.x))&&
(max(u.s.y,u.e.y)>=min(v.s.y,v.e.y))&&
(max(v.s.y,v.e.y)>=min(u.s.y,u.e.y))&&
(mul(v.s,u.e,u.s)*mul(u.e,v.e,u.s)>=0)&&
(mul(u.s,v.e,v.s)*mul(v.e,u.e,v.s)>=0));
}
double dist(double x1,double y1,double x2,double y2)
{
return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
double mian(double x1,double y1,double x2,double y2,double x3,double y3)
{
double d1,d2,d3,p;
d1=dist(x1,y1,x2,y2);
d2=dist(x2,y2,x3,y3);
d3=dist(x1,y1,x3,y3);
p=(d1+d2+d3)/2;
return sqrt(p*(p-d1)*(p-d2)*(p-d3));
}
int main()
{
double x[110],y[110];
int t,n,i;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=0;iscanf("%lf%lf",&x[i],&y[i]);
}
if(n<3){printf("NO\n");continue;}
if(n==3)
{
if((y[1]-y[0])*(x[2]-x[0])==(x[1]-x[0])*(y[2]-y[0]))
{
printf("YES\n");
double d0,d1,d2;
d0=dist(x[1],y[1],x[2],y[2]);
d1=dist(x[0],y[0],x[2],y[2]);
d2=dist(x[1],y[1],x[0],y[0]);
if(d0==d1+d2)printf("BAA\n");
else if(d1==d0+d2)printf("ABA\n");
else printf("AAB\n");
}
else printf("NO\n");
continue;
}
if(n>=4)
{
printf("YES\n");
if((y[1]-y[0])*(x[2]-x[0])==(x[1]-x[0])*(y[2]-y[0]))
{
double d0,d1,d2;
d0=dist(x[1],y[1],x[2],y[2]);
d1=dist(x[0],y[0],x[2],y[2]);
d2=dist(x[1],y[1],x[0],y[0]);
if(d0==d1+d2)printf("BAA");
else if(d1==d0+d2)printf("ABA");
else printf("AAB");
for(i=0;i3;i++)printf("A");
printf("\n");
}
else
{
double s0,s1,s2,s3;
s3=mian(x[0],y[0],x[1],y[1],x[2],y[2]);
s2=mian(x[0],y[0],x[1],y[1],x[3],y[3]);
s1=mian(x[0],y[0],x[3],y[3],x[2],y[2]);
s0=mian(x[3],y[3],x[1],y[1],x[2],y[2]);
if(sgn(s0+s1+s2-s3)==0)printf("AAAB");
else if(sgn(s0+s1-s2+s3)==0)printf("AABA");
else if(sgn(s0-s1+s2+s3)==0)printf("ABAA");
else if(sgn(s0-s1-s2-s3)==0)printf("ABBB");
else
{
point a,b,c,d;
a.x=x[0];
a.y=y[0];
b.x=x[1];
b.y=y[1];
c.x=x[2];
c.y=y[2];
d.x=x[3];
d.y=y[3];
line ab,ac,ad,bc,bd,cd;
ab.e=a;
ab.s=b;
ac.e=a;
ac.s=c;
ad.e=a;
ad.s=d;
bc.e=c;
bc.s=b;
bd.e=d;
bd.s=b;
cd.e=d;
cd.s=c;
if(inter(ab,cd))printf("AABB");
else if(inter(ac,bd))printf("ABAB");
else if(inter(ad,bc))printf("ABBA");
}
for(i=0;i4;i++)printf("A");
printf("\n");
}
}
}
}
I 线段树模板题：
http://blog.csdn.net/lzc504603913/article/details/78072624
G 找规律，比赛时一直卡在这里
同校队伍有用扩展欧几里得过的
贴一个拓展欧几里得算法的介绍
http://blog.csdn.net/zhjchengfeng5/article/details/7786595
下面还有两个大佬的博客
http://www.cnblogs.com/Just–Do–It/p/7583613.html
这个介绍的比较详细，但是感觉还是有点玄，好多很显然
https://www.q-cs.cn/2017/09/23/bounce2017-acm-%E5%8C%97%E4%BA%AC%E7%AB%99%E7%BD%91%E7%BB%9C%E8%B5%9B/