UVa 10691 - Subway (思维 + 区间选点)
题意
每个地铁有一段有效距离,问最少需要多少段地铁。
思路
处理一下每个点的左距离和有距离,问题就变成了区间选点。
但是这个是个圆,所以要都加上2PI来算一遍。
代码
#include <stack>
#include <cstdio>
#include <list>
#include <cassert>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const int MAXN = 1e5 + 10;
const int MOD = 9901;
const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };
int cases = 0;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
int cmp(pdd a, pdd b)
{
return a.Y < b.Y;
}
vector<pdd> arr;
int Solve()
{
int ans = SZ(arr);
sort(arr.begin(), arr.end(), cmp);
int tmpSize = SZ(arr);
for (int i = 0; i < tmpSize; i++) arr.PB(MP(arr[i].X+2*PI, arr[i].Y+2*PI));
for (int i = 0; i < tmpSize; i++)
{
int cnt = 1;
double cur = arr[i].Y;
for (int j = i+1; j < i+tmpSize; j++)
if (arr[j].X-cur > eps)
{
cnt++;
cur = arr[j].Y;
}
ans = min(ans, cnt);
}
return ans;
}
double Dis(double x, double y)
{
return hypot(x, y);
}
int main()
{
//ROP;
int T;
scanf("%d", &T);
while (T--)
{
arr.clear();
int n;
double dis;
scanf("%d%lf", &n, &dis);
for (int i = 0; i < n; i++)
{
double x, y;
scanf("%lf%lf", &x, &y);
if (Dis(x, y) <= dis) continue;
double arc = atan2(y, x);
double add = asin(dis / Dis(x, y));
arr.PB(MP(arc-add, arc+add));
if (arr.back().Y > PI) arr.back().X -= 2*PI, arr.back().Y -= 2*PI;
}
printf("%d\n", Solve());
}
return 0;
}