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ZOJ Problem Set - 2102 Tables

程序员文章站 2022-04-01 13:35:11
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ZOJ Problem Set - 2102
Tables

Time Limit: 2 Seconds      Memory Limit: 65536 KB

Granny likes dropping around in the neighborhood. She is rather aged so she always carries a stick to assist walking. When she comes to a house, she wants to put her stick somewhere, actually anywhere but on the ground, otherwise she has to bend over to pick it up and what a misery it is for her! So she finds some round tables. Err�� so far so good, except the fact that, you know, she is not a physicist. She takes out a PDA and calculates whether the stick will fall to the ground. And she is so proud that the software is written by you, her favorite grandchild.


Input

Input contains multiple test cases. Each case begins with an integer N (0 < N <= 10000) - the number of round tables. N lines follow, each of which contains three integers, xi, yi and ri - the center and radius of the ith round table. Tables do not overlap with each other. The last line contains four integers x1, y1, x2 and y2, where (x1, y1) and (x2, y2) is the two end points of the stick. You may assume the stick is thin and its mass is equally distributed.

A test case with a single 0 signals the end of input, and this test case is not to be processed.


Output

One line for each test case, which contains a single word "STAY" or "FALL". Granny is a reasonable person and she will not bother the program about cases that it might or might not fall. But unfortunately, as mentioned earlier, she is unable to instruct you about physics.


Sample Input

1
0 0 2
0 0 1 1
1
0 0 2
0 0 9 9
2
0 0 2
9 9 2
0 0 9 9
0


Sample Output

STAY
FALL
STAY



Author: WU, Jiazhi
Source: Zhejiang University Local Contest 2004
这题的意思棒子能在桌子上不下落有两种情况1.是当棒子的重心在桌子上 2.当重心不在桌子上时重心的两端都有部分在桌子上关键是第二种情况怎么判断,那么我们可以这样做,我们依据重心把棒子分成两部分,一部分一部分判断,如果其中一个端点和重心都在当前桌子圆心的同侧时,那么最短距离就是其中一个端点到圆心的距离,当不在同侧时,就用叉积来求桌子圆心到棒子的最短距离,在三维空间,两个向量的叉积还是一个向量,并且垂直于另外两个向量,该向量的模为|a|*|b|*sin夹角,坐标公式为:二维空间的向量的叉积,z坐标为0,那么公式简化为a*b=axby-aybx,所以 垂直距离就为axby-aybx / (棒子一部分的长度)
#include <bits/stdc++.h>

using namespace std;

struct node{
    double x,y,r;
}th[10010];
double dis(double x1, double y1, double x2, double y2)
{
    return sqrt((x1-x2)*(x1-x2)*1.0 + (y1-y2)*(y1-y2)*1.0);
}

bool fou(double x1,double y1,double x2,double y2){
    if(x1*x2 + y1*y2 >= 0) return true;
    else return false;
}
int main()
{
    int n;
    while(scanf("%d", &n) != EOF && n){
        for(int  i = 0; i < n; i ++){
            scanf("%lf %lf %lf", &th[i].x, &th[i].y, &th[i].r);
        }
        double x1,x2,y1,y2;
        scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);
        bool zx = false, l = false, r = false;
        double xz = (x1 + x2) / 2.0, yz = (y1 + y2) / 2.0;
        int i;
        for(i = 0; i < n; i ++){
            if(dis(xz, yz, th[i].x, th[i].y) <= th[i].r){
                zx = true;
            }else {
                double xline, yline, xline2, yline2;
                double xpoint, ypoint, xpoint2, ypoint2;
                xline = x1-xz;
                yline = y1-yz;
                xpoint = x1-th[i].x;
                ypoint = y1-th[i].y;

                xline2 = x2-xz;
                yline2 = y2-yz;
                xpoint2 = x2-th[i].x;
                ypoint2 = y2-th[i].y;
                if(fou(xline, yline, xpoint, ypoint) && fou(xz-x1, yz-y1, xz-th[i].x, yz-th[i].y)){
                    double ju = fabs(xline*ypoint - yline*xpoint) / dis(x1, y1, xz,yz);
                    if(ju <= th[i].r){
                        if(x1 == min(x1, x2)) l = true;
                        else r = true;
                    }
                }else {
                    double ju = min(dis(xz, yz, th[i].x, th[i].y), dis(x1,y1, th[i].x, th[i].y));
                    if(ju <= th[i].r) {
                        if(x1 == min(x1, x2)) l = true;
                        else r = true;
                    }
                }
                 if(fou(xline2, yline2, xpoint2, ypoint2) && fou(xz-x2,yz-y2,xz-th[i].x,yz-th[i].y)){
                    double ju = fabs(xline2*ypoint2 - yline2*xpoint2) / dis(x2, y2, xz,yz);
                    if(ju <= th[i].r){
                        if(x2 == min(x1, x2)) l = true;
                        else r = true;
                    }
                }else {
                    double ju = min(dis(xz, yz, th[i].x, th[i].y), dis(x2,y2, th[i].x, th[i].y));
                    if(ju <= th[i].r) {
                        if(x2 == min(x1, x2)) l = true;
                        else r = true;
                    }
                }
            }
            if(zx || (l && r)) break;
        }
        if(i == n) cout << "FALL" << endl;
        else cout << "STAY" <<endl;
    }
    return 0;
}



相关标签: 数学:叉积