题意大概是x轴上方是海洋，海洋上分布着一些岛屿（用点坐标表示），你呢想在岸上安装雷达，使得雷达能扫到岛屿（即在雷达的范围内），求安装雷达的最小数量^-^
Poj-1328 #include<cstdio> #include<algorithm> #include<cmath> const int INF = 1000000000; using namespace std; struct Interval { //interval是区间的意思 double left, right; }; const int max_island = 1000 + 10; Interval the_interval[max_island]; int the_num; double the_r; bool cmp(const Interval& a, const Interval& b) { //排序函数 if (a.right == b.right) return a.left >= b.left; return a.right < b.right; } int main() { int the_case = 1; while (scanf("%d%lf", &the_num, &the_r) == 2) { if (the_num == 0 && the_r == 0) break; double point_x, point_y; bool if_out_r = false; for (int i = 0; i < the_num; i++) { scanf("%lf%lf", &point_x, &point_y); if (point_y > the_r) {//如果离海岸太远，超出雷达半径，我劝你还是放弃吧 if_out_r = true; } else { the_interval[i].left = point_x - sqrt(pow(the_r, 2) - pow(point_y, 2)); the_interval[i].right = point_x + sqrt(pow(the_r, 2) - pow(point_y, 2)); } } if (if_out_r) { printf("Case %d: -1\n", the_case++); } else { sort(the_interval, the_interval + the_num, cmp); double most_left = -INF; //double 我写成int找了很久，最大允许的左边区间点 int the_need_radar = 0; for (int i = 0; i < the_num; i++) { if (the_interval[i].left > most_left) { most_left = the_interval[i].right; the_need_radar++; } } printf("Case %d: %d\n", the_case++, the_need_radar); } } return 0; }总结： 1.防止思路错误——我先前以为是像平移法一样把圆往右移，依次计算，导致我花了很长一段时间来写 2.审题要仔细，特别是那些对变量的描述的词汇，特别注意如数据类型，范围，输出格式······ 3.编程时先写主函数，把框架先写好，遇到阻碍，在写细节。
