POJ_1066_Treasure Hunt_判斷線段相交
阿新 • • 發佈：2018-05-10
i++ under always wan truct eve Go sca 目的
An example is shown below:
POJ_1066_Treasure Hunt_判斷線段相交
Description
Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid of Key-Ops. Using state-of-the-art technology they are able to determine that the lower floor of the pyramid is constructed from a series of straightline walls, which intersect to form numerous enclosed chambers. Currently, no doors exist to allow access to any chamber. This state-of-the-art technology has also pinpointed the location of the treasure room. What these dedicated (and greedy) archeologists want to do is blast doors through the walls to get to the treasure room. However, to minimize the damage to the artwork in the intervening chambers (and stay under their government grant for dynamite) they want to blast through the minimum number of doors. For structural integrity purposes, doors should only be blasted at the midpoint of the wall of the room being entered. You are to write a program which determines this minimum number of doors.An example is shown below:
Input
The input will consist of one case. The first line will be an integer n (0 <= n <= 30) specifying number of interior walls, followed by n lines containing integer endpoints of each wall x1 y1 x2 y2 . The 4 enclosing walls of the pyramid have fixed endpoints at (0,0); (0,100); (100,100) and (100,0) and are not included in the list of walls. The interior walls always span from one exterior wall to another exterior wall and are arranged such that no more than two walls intersect at any point. You may assume that no two given walls coincide. After the listing of the interior walls there will be one final line containing the floating point coordinates of the treasure in the treasure room (guaranteed not to lie on a wall).Output
Print a single line listing the minimum number of doors which need to be created, in the format shown below.Sample Input
7
20 0 37 100
40 0 76 100
85 0 0 75
100 90 0 90
0 71 100 61
0 14 100 38
100 47 47 100
54.5 55.4
Sample Output
Number of doors = 2
來自古文物和古玩博物館（ACM）的考古學家已經飛到埃及去研究大金字塔的大金字塔。
利用最先進的技術，他們能夠確定金字塔的底層是由一系列的直線墻構成的，這些墻與許多封閉的房間相交。目前，沒有任何門可以進入任何一個房間。
這種最先進的技術也精確定位了寶藏室的位置。
這些專門的（和貪婪的）考古學家想要做的是把門從墻上炸開，到達寶藏室。然而，為了盡量減少在中間的房間裏對藝術作品的損害（並在政府的資助下，他們想要通過最少的門數來進行爆炸）。
出於結構完整性的目的，門應該只在進入房間的墻壁的中點被炸開。你要寫一個程序來確定這個最小的門數。
註意是進入房間的墻壁的中點而不是直線墻的中點。
那樣這道題就變得非常簡單了，直接枚舉每個點判斷這個點與寶藏的連線經過多少條線段即可。
然後+1就是答案。
代碼：
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <math.h>
using namespace std;
typedef double f2;
struct Point {
f2 x,y;
Point() {}
Point(f2 x_,f2 y_) :
x(x_),y(y_) {}
Point operator + (const Point &p) const {return Point(x+p.x,y+p.y);}
Point operator - (const Point &p) const {return Point(x-p.x,y-p.y);}
Point operator * (f2 rate) const {return Point(x*rate,y*rate);}
};
struct Line {
Point p,v;
Line() {}
Line(const Point &p_,const Point &v_) :
p(p_),v(v_) {}
};
f2 cross(const Point &p1,const Point &p2) {return p1.x*p2.y-p1.y*p2.x;}
f2 turn(const Point &p1,const Point &p2,const Point &p3) {
return cross(p3-p1,p2-p1);
}
bool judge(const Line &l1,const Line &l2) {
if(turn(l1.p,l1.v,l2.p)*turn(l1.p,l1.v,l2.v)>=0) return 0;
if(turn(l2.p,l2.v,l1.p)*turn(l2.p,l2.v,l1.v)>=0) return 0;
return 1;
}
Line a[50];
Point T;
int n;
int solve(const Line &l) {
int i,re=0;
for(i=1;i<=n;i++) {
if(judge(a[i],l)) re++;
}
return re;
}
int main() {
scanf("%d",&n);
if(!n) {
printf("Number of doors = %d",1); return 0;
}
int i;
int ans=1<<30;
for(i=1;i<=n;i++) {
scanf("%lf%lf%lf%lf",&a[i].p.x,&a[i].p.y,&a[i].v.x,&a[i].v.y);
}
scanf("%lf%lf",&T.x,&T.y);
for(i=1;i<=n;i++) {
Line l=Line(T,a[i].p);
ans=min(ans,solve(l));
l=Line(T,a[i].v);
ans=min(ans,solve(l));
}
printf("Number of doors = %d",ans+1);
}
POJ_1066_Treasure Hunt_判斷線段相交