poj 1066 Treasure Hunt

poj 1066 Treasure Hunt

题意：一个正方形的房间里有很多强，门在每条线段的中，要从一个起点到达最外面问要经过的最少的门。

思路：其实是一个很简单的计算几何题目。把问题转换一下，虽然说门在中点，但是如果要经过这个门，必然要与这条线段相交。这样就变成了判断线段相交的问题。只用枚举，起点到终点(边界上的每一个点)与每条线段的交点个数的最小值。

/* * File: main.cpp * Author: Mi * * Created on 2011年5月11日, 下午8:51 */ #include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> #include <map> #include <vector> #include <stdlib.h> #define N 105 #define eps 1e-8 using namespace std; /* * */ struct Point { double x,y; }start,p[N]; struct Line { Point st,ed; }l[N]; int dblcmp(double d) { if(fabs(d)<eps) return 0; return d>0?1:-1; } int xyCmp(double p,double mini,double maxi) { return dblcmp(p-mini)*dblcmp(p-maxi); } int betweenCmp(Point a,Point b,Point c) { if(fabs(b.x-c.x)>fabs(b.y-c.y)) return xyCmp(a.x,min(b.x,c.x),max(b.x,c.x)); else return xyCmp(a.y,min(b.y,c.y),max(b.y,c.y)); } double det(double x1,double y1,double x2,double y2)//area2 { return x1*y2-x2*y1; } double cross(Point a,Point b,Point c) { return det(b.x-a.x,b.y-a.y,c.x-a.x,c.y-a.y); } int segcross(Point a,Point b,Point c,Point d) { double s1,s2,s3,s4; int d1,d2,d3,d4; d1=dblcmp(s1=cross(a,b,c)); d2=dblcmp(s2=cross(a,b,d)); d3=dblcmp(s3=cross(c,d,a)); d4=dblcmp(s4=cross(c,d,b)); if((d1^d2)==-2&&(d3^d4)==-2)//规范相交 return 1; if(d1==0&&betweenCmp(c,a,b)<=0||d2==0&&betweenCmp(d,a,b)<=0|| d3==0&&betweenCmp(a,c,d)<=0||d4==0&&betweenCmp(b,c,d)<=0) return 2;//非规范相交 return 0; } int main(int argc, char** argv) { int n,cnt=0,ans=999999999; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%lf%lf%lf%lf",&l[i].st.x,&l[i].st.y,&l[i].ed.x,&l[i].ed.y); p[cnt++]=l[i].st; p[cnt++]=l[i].ed; } scanf("%lf%lf",&start.x,&start.y); for(int i=0;i<cnt;i++) { int count=0; for(int j=0;j<n;j++) { if(segcross(start,p[i],l[j].st,l[j].ed)) count++; } ans=min(count,ans); } if(n==0) ans=1; printf("Number of doors = %d/n",ans); return 0; }