Description Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d. We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.  Figure A Sample Input of Radar Installations Input The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases. The input is terminated by a line containing pair of zeros Output For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case. Sample Input 3 21 2-3 12 11 20 20 0 Sample Output Case 1: 2Case 2: 1 Source |
海上的每一个岛到陆地上最远距离d有两个点，当雷达在这两个点所确定的区域时，这个岛就在雷达所在的范围内。所以就将这个问题化为贪心的区间覆盖问题。将所有岛的区间左端点进行升序排列，每次使右端点所能包含的线段数目最大（答案为统计左右不重合的线段的数目）。
#include#include #include #include #define MAX_N 1010#define MAX(a, b) ((a > b)? a: b)#define MIN(a, b) ((a < b)? a: b)using namespace std;const int INF = 1e8;struct node{ double l, r;}pon[MAX_N];bool cmp(node x, node y) { return x.l < y.l;}int main() { int n, d; double x, y; int cnt = 0; while (scanf("%d%d", &n, &d), n||d) { int ans = 0; bool flag = false; for (int i = 0; i < n; i++) { scanf("%lf%lf", &x, &y); if (y > d) { flag = true; } pon[i].l = x - sqrt(d*d - y*y); pon[i].r = x + sqrt(d*d - y*y); } if (flag) { printf("Case %d: -1\n", ++cnt); continue; } sort(pon, pon + n, cmp); double t = pon[0].r; for (int i = 1; i < n; i++) { if(pon[i].l > t) { t = pon[i].r; ans++; } //如果当前的右端点大于pon[i].r，则当前的右端点应等于pon[i].r else if(pon[i].r < t) { t = pon[i].r; } } printf("Case %d: %d\n", ++cnt, ans + 1); } return 0;}