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Coneology
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 3289 | Accepted: 586 |
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 dominateother 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 Ri, 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 Ri, xi, yi separated by spaces, where (xi, yi) 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
这题并不难，和多校的某道题稍微思想上有点像。
问不包含于其他圆内部的圆的数目。
取所有的左端点和右端点，排序后一次取，对于在内部的圆，其外部的圆一定只取到了左端点，那么我们对于当前只取到左端点的圆，只需要判断这个圆的纵坐标是否在那些只取到左端点的圆的纵坐标之间。因为不会相交，所以可以这样搞。
1 /**
2 * code generated by JHelper
3 * More info: https://github.com/AlexeyDmitriev/JHelper
4 * @author xyiyy @https://github.com/xyiyy
5 */
6
7 #include <iostream>
8 #include <fstream>
9
10 //#####################
11 //Author:fraud
12 //Blog: http://www.cnblogs.com/fraud/
13 //#####################
14 //#pragma comment(linker, "/STACK:102400000,102400000")
15 #include <iostream>
16 #include <sstream>
17 #include <ios>
18 #include <iomanip>
19 #include <functional>
20 #include <algorithm>
21 #include <vector>
22 #include <string>
23 #include <list>
24 #include <queue>
25 #include <deque>
26 #include <stack>
27 #include <set>
28 #include <map>
29 #include <cstdio>
30 #include <cstdlib>
31 #include <cmath>
32 #include <cstring>
33 #include <climits>
34 #include <cctype>
35
36 using namespace std;
37 #define mp(X, Y) make_pair(X,Y)
38 #define pb(X) push_back(X)
39 #define rep(X, N) for(int X=0;X<N;X++)
40 #define IT iterator
41 #define ALL(X) (X).begin(),(X).end()
42
43 double r[40010], x[40010], y[40010];
44 vector<pair<double, int> > vec;
45 set<pair<double, int> > s;
46 bool mark[40010];
47 vector<int> ans;
48
49 bool contain(int a, int b) {
50 return ((x[a] - x[b]) * (x[a] - x[b]) + (y[a] - y[b]) * (y[a] - y[b]) <= r[b] * r[b]);
51 }
52
53 class poj2932 {
54 public:
55 void solve(std::istream &in, std::ostream &out) {
56 int n;
57 while (in >> n) {
58 vec.clear();
59 s.clear();
60 ans.clear();
61 rep(i, n)in >> r[i] >> x[i] >> y[i];
62 rep(i, n) {
63 vec.pb(mp(x[i] - r[i], i));
64 vec.pb(mp(x[i] + r[i], i + n));
65 }
66 sort(ALL(vec));
67 int num = 0;
68 rep(i, vec.size()) {
69 int j = vec[i].second;
70 if (j < n) {
71 set<pair<double, int> >::IT it = s.lower_bound(mp(y[j], j));
72 if (it != s.end() && contain(j, it->second))continue;
73 if (it != s.begin() && contain(j, (--it)->second))continue;
74 s.insert(mp(y[j], j));
75 mark[j + 1] = 1;
76 num++;
77 } else {
78 s.erase(mp(y[j - n], j - n));
79 }
80 }
81 out << num << endl;
82 rep(i, n) {
83 if (mark[i + 1])out << i + 1 << " ";
84 }
85 out << endl;
86 }
87 }
88 };
89
90 int main() {
91 std::ios::sync_with_stdio(false);
92 std::cin.tie(0);
93 poj2932 solver;
94 std::istream &in(std::cin);
95 std::ostream &out(std::cout);
96 solver.solve(in, out);
97 return 0;
98 }
代码君