UVALive 7461 - Separating Pebbles

传送门

题目大意：
给一个数N，然后每行3个数，前两个数十坐标，后面的一个数区分是 ‘+’ 还是’o’，0代表的是’o’，1代表的是’+’.然后判断是不是能够有一条线把这两部分分开也就是线的两边都是一样的标识符。如果能输出1，否则输出 0.

解题思路：
其实，我们可以将’o’代表的所有的点构成一个凸包，将’+’代表的所有的点构成一个凸包，判断这两个凸包是不是有相交就行了。

My Code：

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
using namespace std;
///基础点和向量运算
struct Point
{double x,y;Point(double x=0,double y=0):x(x),y(y) {}
};
typedef Point Vector;
Vector operator + (Vector A,Vector B)
{return Vector(A.x+B.x,A.y+B.y);
}
Vector operator - (Vector A,Vector B)
{return Vector(A.x-B.x,A.y-B.y);
}
Vector operator * (Vector A,double p)
{return Vector(A.x*p,A.y*p);
}
Vector operator / (Vector A,double p)
{return Vector(A.x/p,A.y/p);
}
bool operator <(const Point& a, const Point& b)
{return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
const double eps=1e-10;
int dcmp(double x)///判断正负，或者等于0
{if(fabs(x)<eps)return 0;else return x<0?-1:1;
}
bool operator==(const Point& a,const Point &b)
{return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Vector A, Vector B)
{return A.x*B.x+A.y*B.y;   ///点积
}
double Length(Vector A)
{return sqrt(Dot(A,A));   ///OA长
}
double Angle(Vector A,Vector B)
{return acos(Dot(A,B)/Length(A)/Length(B));   ///OA和OB的夹角
}
double Cross(Vector A,Vector B)
{return A.x*B.y-A.y*B.x;   ///叉积
}
double Area2(Point A,Point B,Point C)
{return Cross(B-A,C-A);   ///三角形面积
}
Vector Rotate(Vector A,double rad)///rad为弧度，旋转rad度
{return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
Vector Normal(Vector A)///A的单位法向量，A不能为零向量
{double L=Length(A);return Vector(-A.y/L,A.x/L);
}///点和直线
///P+tv表示一条直线，P为点，tv为方向向量
Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)///求直线交点，确保存在交点，即Cross(v,w)非0
{Vector u=P-Q;double t=Cross(w,u)/Cross(v,w);return P+v*t;
}
double DistanceToLine(Point P,Point A,Point B)///P点到直线AB的距离
{Vector v1=B-A,v2=P-A;return fabs(Cross(v1,v2)/Length(v1));
}
double DistanceToSegment(Point P,Point A,Point B)///点P到线段AB的距离
{if(A==B)return Length(P-A);Vector v1=B-A,v2=P-A,v3=P-B;if(dcmp(Dot(v1,v2))<0)return Length(v2);else if(dcmp(Dot(v1,v3))>0)return Length(v3);else return fabs(Cross(v1,v2)/Length(v1));
}
Point GetLineProjection(Point P,Point A,Point B)///点在直线上的投影
{Vector v=B-A;return A+v*(Dot(v,P-A)/Dot(v,v));
}
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)///判断线段相交,不在端点相交
{double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1),c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}
bool OnSegment(Point p,Point a1,Point a2)///判断点是否在线段上（不包括端点）
{return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}///多边型
double ConvexPolygonArea(Point* p,int n)///多边形面积,，点按顺序
{double area=0;for(int i=1; i<n-1; i++)area+=Cross(p[i]-p[0],p[i+1]-p[0]);return area/2;
}const int maxn=1e3+10;
int ConvexHull(Point *p,Point *ch,int n)///求凸包
{sort(p,p+n);int i,m=0,k;for(i=0; i<n; i++){while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;ch[m++]=p[i];}k=m;for(i=n-2; i>=0; i--){while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;ch[m++]=p[i];}if(n>1)m--;return m;
}///if(ConvexHullIntersection(ch1,t1,ch2,t2)&&ConvexHullIntersection(ch2,t2,ch1,t1))printf("YES\n");else printf("NO\n");
bool ConvexHullIntersection(Point *ch1,int t1,Point *ch2,int t2)///判断凸包是否相交
{double angle[maxn],x;int i,j,k,m;if(t1==1)return true;if(t1==2){for(i=0; i<t2; i++){k=dcmp(Cross(ch1[1]-ch1[0],ch2[i]-ch1[0]));if(k==0&&Dot(ch1[1]-ch1[0],ch2[i]-ch1[0])>0){if(Length(ch2[i]-ch1[0])<Length(ch1[1]-ch1[0]))break;}}if(i<t2)return false;if(t2==2&&SegmentProperIntersection(ch1[0],ch1[1],ch2[0],ch2[1]))return false;return true;}angle[0]=0;for(i=2; i<t1; i++)angle[i-1]=Angle(ch1[1]-ch1[0],ch1[i]-ch1[0]);for(i=0; i<t2; i++){j=dcmp(Cross(ch1[1]-ch1[0],ch2[i]-ch1[0]));if(j<0||(j==0&&Dot(ch1[1]-ch1[0],ch2[i]-ch1[0])<0))continue;j=dcmp(Cross(ch1[t1-1]-ch1[0],ch2[i]-ch1[0]));if(j>0||(j==0&&Dot(ch1[t1-1]-ch1[0],ch2[i]-ch1[0])<0))continue;x=Angle(ch1[1]-ch1[0],ch2[i]-ch1[0]);m=lower_bound(angle,angle+t1-1,x)-angle;if(m==0)j=0;else j=m-1;k=dcmp(Cross(ch1[j+1]-ch2[i],ch1[j+2]-ch2[i]));if(k>=0)break;}if(i<t2)return false;return true;
}Point p1[maxn],ch1[maxn],p2[maxn],ch2[maxn], p[maxn];
int n,m,t1,t2;
int main()
{int T;cin>>T;while(T--){int sum;cin>>sum;int i,j,k, n = 0, m = 0, op;for(i=0; i<sum; i++){scanf("%lf%lf%d",&p[i].x,&p[i].y,&op);if(op==0){p1[n].x = p[i].x;p1[n++].y = p[i].y;}else{p2[m].x = p[i].x;p2[m++].y = p[i].y;}}t1=ConvexHull(p1,ch1,n);t2=ConvexHull(p2,ch2,m);if(ConvexHullIntersection(ch1,t1,ch2,t2)&&ConvexHullIntersection(ch2,t2,ch1,t1))puts("1");elseputs("0");}return 0;
}

UVALive 7461 - Separating Pebbles相关推荐

DP UVALive 6506 Padovan Sequence
题目传送门 /*题意:两行数字,相邻列一上一下,或者隔一列两行都可以,从左到右选择数字使和最大DP:状态转移方程:dp[i][j] = max (dp[i][j], dp[1-i][j-1] + a[ ...
The UVALIVE 7716 二维区间第k小
The UVALIVE 7716 二维区间第k小 /** 题意:给一个n * n的矩阵,有q个查询每次查询r,c,s,k表示已(r,c)为右上角大小为s的正方形中第k小的元素n <= 250 ...
UVALive 8513 lovers 2017 西安区域赛 B 贪心+multiset
UVALive 8513 有2种人,每个人有自己的权值$A_i$ $B_i$ 当$A_i + B_i >=K$时两个人可以配对问最多多少人可以配对解法 : 把$/{ A_i /}$ 排序 ...
训练指南 UVALive - 3713 （2-SAT）
layout: post title: 训练指南 UVALive - 3713 (2-SAT) author: "luowentaoaa" catalog: true mathja ...
逆序数 UVALive 6508 Permutation Graphs
题目传送门 1 /* 2 题意:给了两行的数字,相同的数字连线,问中间交点的个数 3 逆序数:第一行保存每个数字的位置,第二行保存该数字在第一行的位置,接下来就是对它求逆序数 4 用归并排序或线段树求 ...
Infinite Fraction Path UVALive - 8207
Infinite Fraction Path UVALive - 8207 题意: 给你n个数,每个数在0到9之间,每个数的下标一次是0~n-1,然后他所能走到的数为(i^2+1)%n,i为他本身的下 ...
F - Heron and His Triangle UVALive - 8206
F - Heron and His Triangle UVALive - 8206 题意: 给你应该n,然后求一个最小的t,问长度为t-1,t,t+1所组成的三角形的面积为整数,t>=n 题解: ...
Tree UVALive - 8212
Tree UVALive - 8212 题意: 有n个点,k个颜色,每个点都要被染色,相同颜色之间的边算是被该颜色覆盖,问有多少边被所有颜色覆盖题解: 题目给的是无根树,我们可以将1默认为根然后求所 ...
Rabbits UVALive - 8211
Rabbits UVALive - 8211 题意: n个兔子的位置,兔子每次可以跳到两个兔子之间,问最多可以跳多少下? 题解: 求出所有相邻两数的间隔,然后减去最小间隔就是答案代码: #inclu ...
Little Boxes UVALive - 8209
Little Boxes UVALive - 8209 题意: 给你四个数,输出四个数之和,四个数小于等于262之内题解: 这...这..水题 unsigned int 0-4294967295 ( ...

UVALive 7461 - Separating Pebbles

UVALive 7461 - Separating Pebbles相关推荐

最新文章

热门文章