POJ-2932 Coneology[平面扫描]

Posted 2020-10-09 zprhhs

Coneology

Time Limit: 5000MS

Description

A student named Round Square loved to play with cones. He would arrange cones with different base radii arbitrarily on the floor and would admire the intrinsic beauty of the arrangement. The student even began theorizing about how some cones dominate other cones: a cone A dominates another cone B when cone B is completely within the cone A. Furthermore, he noted that there are some cones that not only dominate others, but are themselves dominated, thus creating complex domination relations. After studying the intricate relations of the cones in more depth, the student reached an important conclusion: there exist some cones, all-powerful cones, that have unique properties: an all-powerful cone is not dominated by any other cone. The student became so impressed by the mightiness of the all-powerful cones that he decided to worship these all-powerful cones.

Unfortunately, after having arranged a huge number of cones and having worked hard on developing this grandiose cone theory, the student become quite confused with all these cones, and he now fears that he might worship the wrong cones (what if there is an evil cone that tries to trick the student into worshiping it?). You need to help this student by finding the cones he should worship.

Input

The input le specifies an arrangement of the cones. There are in total N cones (1 ≤ N ≤ 40000). Cone i has radius and height equal to R_i, i = 1 … N. Each cone is hollow on the inside and has no base, so it can be placed over another cone with smaller radius. No two cones touch.

The first line of the input contains the integer N. The next N lines each contain three real numbers R_i, x_i, y_i separated by spaces, where (x_i, y_i) are the coordinates of the center of the base of cone i.

Output

The first line of the output le should contain the number of cones that the student should worship. The second line contains the indices of the cones that the student should worship in increasing order. Two consecutive numbers should be separated by a single space.

Sample Input

5
1 0 -2
3 0 3
10 0 0
1 0 1.5
10 50 50

Sample Output

2
3 5

题意：给出n个不相交的圆，求出最外层的圆，也就是不在其他圆内部的圆。

思路：

这是挑战上平面扫描的一个例题。扫描线的思想比较好理解，就是一条直线在不断的扫描的过程中更新信息。

对于这道题而言，直线平行于y轴从左到扫描，每次到达一个圆的最左边，找的和这个圆最近两个圆的y坐标，检查这两个圆和当前的位置。每次从最左边开始，保证的最外面的圆是首先进入。

#include "stdio.h"
#include "math.h"
#include "vector"
#include "set"
#include "algorithm"
using namespace std;
const int maxn = 4e4 + 10;
int N;
struct Cri {
    double x, y, r;
};
Cri c[maxn];
bool inside(int i, int j) {
    double dx = c[i].x - c[j].x, dy = c[i].y - c[j].y;
    return dx*dx+dy*dy <= c[j].r*c[j].r;
}
vector<int> res;
vector<pair<double, int>> events;
set<pair<double, int>> outs;
int main(int argc, char const *argv[])
{
    scanf("%d", &N);
    for (int i = 0; i < N; i++) {
        scanf("%lf %lf %lf", &c[i].r, &c[i].x, &c[i].y);
        events.push_back(make_pair(c[i].x - c[i].r, i));
        events.push_back(make_pair(c[i].x + c[i].r, i + N));
    }
    sort(events.begin(),events.end());
    for (int i = 0; i < events.size(); i++) {    
        int id = events[i].second%N;
        if (events[i].second < N) {
            set<pair<double, int>>::iterator it = outs.lower_bound(make_pair(c[id].y, id));
            if (it != outs.end() && inside(id, it->second)) continue;
            if (it != outs.begin() && inside(id, (--it)->second)) continue;
            res.push_back(id);
            outs.insert(make_pair(c[id].y, id));
         }
         else outs.erase(make_pair(c[id].y, id));
    }
    sort(res.begin(), res.end());
    printf("%d\n", res.size());
    for (int i = 0; i < res.size(); i++) {
        printf("%d%c", res[i]+1, i+1==res.size()? ‘\n‘:‘ ‘);
    }
    return 0;
}