ZOJ1648 Circuit Board 2017-04-18 20:31 34人阅读 评论(0) 收藏

Circuit Board

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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On the circuit board, there are lots of circuit paths. We know the basic constrain is that no two path cross each other, for otherwise the board will be burned.



Now given a circuit diagram, your task is to lookup if there are some crossed paths. If not find, print "ok!", otherwise "burned!" in one line.



A circuit path is defined as a line segment on a plane with two endpoints p1(x1,y1) and p2(x2,y2).



You may assume that no two paths will cross each other at any of their endpoints.

Input



The input consists of several test cases. For each case, the first line contains an integer n(<=2000), the number of paths, then followed by n lines each with four float numbers x1, y1, x2, y2.

Output



If there are two paths crossing each other, output "burned!" in one line; otherwise output "ok!" in one line.

Sample Input

1

0 0 1 1

2

0 0 1 1

0 1 1 0

Sample Output

ok!

burned!

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题目的意思是给出n条线段，问是否有线段相交

两两之间暴力判，判可以用模板

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <set>
#include <stack>
#include <map>
#include <functional>
#include <bitset>
#include <string>

using namespace std;

#define LL long long
#define INF 0x3f3f3f3f

struct Point
{
    double x;
    double y;
};
typedef struct Point point;
struct line
{
    point p1,p2;
};

double multi(point p0,point p1,point p2)
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
bool is(point s1,point e1,point s2,point e2)
{
    return (max(s1.x,e1.x)>=min(s2.x,e2.x))&&
           (max(s2.x,e2.x)>=min(s1.x,e1.x))&&
           (max(s1.y,e1.y)>=min(s2.y,e2.y))&&
           (max(s2.y,e2.y)>=min(s1.y,e1.y))&&
           (multi(s1,s2,e1)*multi(s1,e1,e2)>0)&&
           (multi(s2,s1,e2)*multi(s2,e2,e1)>0);
}

int main()
{
    int n;
    line L[2005];
    while(~scanf("%d",&n))
    {
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf%lf%lf",&L[i].p1.x,&L[i].p1.y,&L[i].p2.x,&L[i].p2.y);
        }
        int flag=0;
        for(int i=0; i<n; i++)
        {
             for(int j=0; j<n; j++)
            {
                if(i==j)
                    continue;
                if(is(L[i].p1,L[i].p2,L[j].p1,L[j].p2))
                {
                    flag=1;
                    break;
                }
            }
            if(flag)
                break;
        }
        if(flag)
            printf("burned!\n");
        else
            printf("ok!\n");
    }
    return 0;
}

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Author: ZHOU, Feng

Source: ZOJ Monthly, September 2003