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

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 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：

#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：