UVALive 2519 Radar Installation 雷达扫描 区间选点问题
题意:在坐标轴中给出n个岛屿的坐标,以及雷达的扫描距离,要求在y=0线上放尽量少的雷达能够覆盖全部岛屿。
很明显的区间选点问题。
代码:
/* * Author: illuz <iilluzen[at]gmail.com> * Blog: http://blog.csdn.net/hcbbt * File: l2911.cpp * Create Date: 2013-09-09 20:51:05 * Descripton: */ #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int MAXN = 1010; int n, r, cas = 1; struct Point { double lhs, rhs; } a[MAXN]; bool cmp(Point a, Point b) { if (a.rhs != b.rhs) return a.rhs < b.rhs; return a.lhs > b.lhs; } int main() { int x, y; bool flag; while (scanf("%d%d", &n, &r) && (n || r)) { flag = true; memset(a, 0, sizeof(a)); for (int i = 0; i < n; i++) { scanf("%d%d", &x, &y); if (flag) { if (y > r) { flag = false; continue; } double t = sqrt((double)r * r - y * y); a[i].lhs = x - t; a[i].rhs = x + t; } } printf("Case %d: ", cas++); if (!flag) printf("-1\n"); else { sort(a, a + n, cmp); // for (int i = 0; i < n; i++) // printf("%lf %lf\n", a[i].lhs, a[i].rhs); int ans = 1; double start = a[0].rhs; for (int i = 1; i < n; i++) { if (a[i].lhs - start <= 1e-4) continue; ans++; start = a[i].rhs; } printf("%d\n", ans); } } return 0; }