poj 2653 Pick-up sticks(几何线段相交判断)

Description

Input

Output

For each input case, print one line of output listing the top sticks in the format given in the sample. The top sticks should be listed in order in which they were thrown. 

The picture to the right below illustrates the first case from input.

Sample Input

5
1 1 4 2
2 3 3 1
1 -2.0 8 4
1 4 8 2
3 3 6 -2.0
3
0 0 1 1
1 0 2 1
2 0 3 1
0

Sample Output

Top sticks: 2, 4, 5.
Top sticks: 1, 2, 3.

Hint

Source

题意：按顺序给出一些木棍，输出在最上面的木棍标号

#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstdio>
#include<vector>
#define N 100010
#define inf 99999999999.0

using namespace std;

struct Point {
    double x,y;
} ;

struct Line {
    Point p1,p2;
} L[N];

int n,num;
bool vis[N];

Point addPoint(double x,double y) {
    Point it;
    it.x=x;
    it.y=y;
    return it;
}

Line addLine(double x1,double y1,double x2,double y2) {
    Point a=addPoint(x1,y1);
    Point b=addPoint(x2,y2);
    Line it;
    it.p1=a,it.p2=b;
    return it;
}

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);
}

///判断是否相交
bool is_inter(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 ans[1010];

int main() {
    //freopen("test.in","r",stdin);
    while(cin>>n&&n) {
        double x,y,_x,_y;
        num=0;
        memset(vis,0,sizeof vis);
        for(int i=0; i<n; i++) {
            scanf("%lf%lf%lf%lf",&x,&y,&_x,&_y);
            L[i]=addLine(x,y,_x,_y);
        }
        for(int i=0; i<n; i++) {
            for(int j=i+1; j<n; j++) {
                if(is_inter(L[i].p1,L[i].p2,L[j].p1,L[j].p2)) {
                    vis[i]=1;
                    break;
                }
            }
        }
        int k=0;
        for(int i=0; i<n; i++) {
            if(!vis[i]) {
                ans[k++]=i+1;
            }
        }
        printf("Top sticks: ");
        for(int i=0; i<k-1; i++)
            printf("%d, ",ans[i]);
        printf("%d.\n",ans[k-1]);
    }
    return 0;
}

poj 2653 Pick-up sticks(几何线段相交判断)

原文：http://blog.csdn.net/acm_baihuzi/article/details/47301881