题目:uva10382 - Watering Grass(区间覆盖变形)
题目大意:要给一片草坪浇水,给定草坪的长度和宽度,给出每个喷头的圆心C和喷水的半径R,问最少要几个喷头可以给整片草坪都浇上水。
解题思路:区间覆盖问题的变形,因为草坪有宽度W,所以这个每个喷头的有效范围是[C- sqrt(R* R - 0.25 * W * W , C + sqrt (R*R - 0.25 * W *W)].
代码:
#include <stdio.h> #include <algorithm> #include <math.h> #include <string.h> using namespace std; const int N = 10005; int n, l, w; int visit[N]; double Max (const double a, const double b) { return a > b ? a: b; } struct Len { double L, R; }len[N]; int cmp (const Len &a, const Len &b) { return a.L < b.L;} void handle (int c, int r, int i) { double r1 = r; double c1 = c; double w1 = w; double temp = sqrt (r1 * r1 - 0.25 * w1 * w1); len[i].L = c - temp; len[i].R = c + temp; } int solve () { memset (visit, 0, sizeof (visit)); if (len[0].L > 0) return -1; double s = 0; double ll = -1; int m = -1; for (int i = 0; i < n; i++) { if (len[i].L <= s) { if (ll < len[i].R) { visit[i] = 1; if (m >= 0) visit[m] = 0; m = i; ll = Max (ll, len[i].R); } } else { if (s == ll) return -1; s = ll; i--; m = -1; } // printf ("%f\n", ll); } // printf ("%f\n", ll); if (ll < l) return -1; int count = 0; for (int i = 0; i < n; i++) if (visit[i]) count++; return count; } int main () { int c, r; while (scanf ("%d%d%d", &n, &l, &w) != EOF) { for (int i = 0; i < n; i++) { scanf ("%d%d", &c, &r); handle (c, r, i); } sort (len, len + n, cmp); /* for (int i = 0; i < n; i++) printf ("%f %f\n", len[i].L, len[i].R);*/ printf ("%d\n", solve()); } return 0; }
时间: 2024-10-16 08:06:54