POJ 1418 Viva Confetti(极角扫描)

题意:

$给定N\le 100个圆,按输入顺序1个1个放,问最后能看到哪些圆$

分析:

$这个题的做法来自于题解了,感觉我也是似懂非懂$
$考虑对于每个圆来说与其他圆的相交部分,是一段圆弧,极角表示法$
$极角排序后,对于任意2个交点的每段圆弧,求它们的中点$
$之后向内向外偏移一点,求上面那个第一个盖住它的就是可以看见的$
$由于是从下往上找的,所以最后留下的就是都可以看见的$
$时间复杂度O(n^3logn)$

$代码:$

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//
// Created by TaoSama on 2016-08-23
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#include <complex>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
typedef complex<double> P;
const double EPS = 5e-13, PI = acos(-1);
int n;
double normalize(double r) {
if(r < 0.0) r += 2 * PI;
if(r >= 2 * PI) r -= 2 * PI;
return r;
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d", &n) == 1 && n) {
vector<P> ps(n);
vector<double> rs(n);
for(int i = 0; i < n; ++i) {
double x, y, r;
scanf("%lf%lf%lf", &x, &y, &r);
ps[i] = P(x, y);
rs[i] = r;
}
vector<int> visible(n, 0);
for(int i = 0; i < n; ++i) {
vector<double> angles;
angles.push_back(0);
angles.push_back(2 * PI);
for(int j = 0; j < n; ++j) {
if(i == j) continue;
P v = ps[j] - ps[i];
double a = abs(v);
double b = rs[i];
double c = rs[j];
if(a + b < c || a + c < b || b + c < a) continue;
double angle = arg(v);
double delta = acos((a * a + b * b - c * c) / (2 * a * b));
angles.push_back(normalize(angle - delta));
angles.push_back(normalize(angle + delta));
}
sort(angles.begin(), angles.end());
for(int j = 0; j + 1 < angles.size(); ++j) {
double theta = (angles[j] + angles[j + 1]) / 2;
for(int d = -1; d <= 1; d += 2) {
P o = ps[i] + polar(rs[i] + d * EPS, theta);
for(int k = n - 1; ~k; --k) {
if(abs(ps[k] - o) < rs[k]) {
visible[k] = true;
break;
}
}
}
}
}
int ans = count(visible.begin(), visible.end(), true);
printf("%d\n", ans);
}
return 0;
}


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