Day3-C-Radar Installation POJ1328
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
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
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 2 1 2 -3 1 2 1 1 2 0 2 0 0
Sample Output
Case 1: 2 Case 2: 1
简述:每个island与X轴都有最多2个交点,求最少点满足与所有区间相交
思路:区间选点问题,将每个区间右边递增排序后寻找即可,代码如下:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#define sqr(x) ((x)*(x)) const int maxm = 1005; struct Node { double l, r; bool operator< (const Node &a) const { return r < a.r; } } Nodes[maxm]; int d, n, sum, kase = 0; int main() { while(scanf("%d%d", &n, &d) && n) { printf("Case %d: ", ++kase); bool flag = true; sum = 1; for (int i = 0; i < n; ++i) { double tx, ty, tmp; scanf("%lf%lf", &tx, &ty); //x = tx -+ sqrt(d^2 - y0 ^2 ) if(d < ty) { flag = false; sum = -1; } tmp = sqrt(sqr(d) - sqr(ty)); Nodes[i].l = tx - tmp, Nodes[i].r = tx + tmp; } if(flag) { sort(Nodes, Nodes + n); double maxr = Nodes[0].r; for (int i = 1; i < n; ++i) { if(maxr < Nodes[i].l) { maxr = Nodes[i].r; ++sum; } } } printf("%d\n", sum); } return 0; }
注意在判断ty>d的时候不能提前退出,要读取完
补:
在区间选点问题上,要右端点进行排序,因为要找一个现有区间的公共点,若是左端点,会出现漏解的情况,例如: