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poj 3304 Segments

2016-05-05 16:14 190 查看

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13156		Accepted: 4207

Description

Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least one point in common.

Input

Input begins with a number T showing the number of test cases and then, T test cases follow. Each test case begins with a line containing a positive integer n ≤ 100 showing the number of segments. After that, n lines containing
four real numbers x1 y1 x2 y2 follow, in which (x1, y1) and (x2, y2) are the coordinates of the
two endpoints for one of the segments.

Output

For each test case, your program must output "Yes!", if a line with desired property exists and must output "No!" otherwise. You must assume that two floating point numbers a and b are equal if |a - b| < 10-8.

Sample Input

3
2
1.0 2.0 3.0 4.0
4.0 5.0 6.0 7.0
3
0.0 0.0 0.0 1.0
0.0 1.0 0.0 2.0
1.0 1.0 2.0 1.0
3
0.0 0.0 0.0 1.0
0.0 2.0 0.0 3.0
1.0 1.0 2.0 1.0

Sample Output

Yes!
Yes!
No!

Source

Amirkabir University of Technology Local Contest 2006

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题意：

给定一些线段，判断存在不存在一条直线2，使任意两条线段在直线2的投影都有公共点

思路：

假如存在一条直线1，可以穿过所有的线段，那么做这条直线1的垂线就是所求直线2(枚举任意两线段的端点，看其他线段的点是否都在这条直线两侧，叉积来做)

代码：

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#define pi acos(-1.0)
#define maxn (100 + 50)
#define mod 1000000007
#define Lowbit(x) (x & (-x))
using namespace std;
typedef long long int LLI;

struct Point {
double x;
double y;
Point() {}
Point(double xx,double yy) {
x = xx,y = yy;
}
Point operator-(const Point &b)const {
return Point(x - b.x,y - b.y);
}
double operator*(const Point &b)const {
return x * b.x + y * b.y;
}
double operator^(const Point &b)const {
return x * b.y - y * b.x;
}
};

struct Line {
Point Start;
Point End;
Line() {}
Line(Point s,Point e) {
Start = s;
End = e;
}
} line[maxn * 2];

double Xmult(Point p0,Point p1,Point p2) {
return (p1 - p0) ^ (p2 - p0);
}
bool Judge(Point a,Point b,int n) {
if(abs(a.x - b.x) < 1e-8 && abs(a.y - b.y) < 1e-8)  return false;
for(int i = 1; i <= n; i ++) {
if(Xmult(line[i].Start,a,b) * Xmult(line[i].End,a,b) > 1e-8)
return false;
}
return true;
}
bool Fun(int n) {
for(int i = 1; i <= n; i ++) {
for(int j = i + 1; j <= n; j ++) {
if(Judge(line[i].Start,line[j].End,n))    return true;
if(Judge(line[i].End,line[j].End,n))      return true;
if(Judge(line[i].Start,line[j].Start,n))  return true;
if(Judge(line[i].End,line[j].Start,n))    return true;
}
}
return false;
}
int main() {
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
int t;
scanf("%d",&t);
while(t --) {
int n;
scanf("%d",&n);
for(int i = 1; i <= n; i ++) {
double x1,x2,y1,y2;
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
line[i] = Line(Point(x1,y1),Point(x2,y2));
}
if(n == 1 || Fun(n) == true)    printf("Yes!\n");(n == 1 时，不能进入循环，需特判)
else                            printf("No!\n");
}
return 0;
}

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