poj1328Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 76972		Accepted: 17240

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 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

题意：在笔直的航线附近有许多岛屿，给定岛屿的坐标，雷达的半径，求至少多少雷达才可以覆盖所有的岛屿，若无法覆盖，输出-1，若行，输出雷达数量（x轴当做航向）
思路：先以每一个点为中心画弧，与x轴相交两点成一块区域，则在区域内任意一点为圆心画圆都可以覆盖那个点，将所有的区间按照每个区间最左端从小到大排序
     从起始区间开始，1：若上一个区间的最右端小于当前区间最左端，上一个雷达将无法覆盖当前区间，则需要换一个新雷达来覆盖当前的区间
                  2：若1条件不成立，即上一个区间最右端大于当前区间最左端，雷达能覆盖当前区间，就需要取两个区间的公共部分，即比较两个区间最右端大小；那个小取哪个
AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<vector>
#include<math.h>

using namespace std;
struct section {
    double left;
    double right;
    //bool operator < (const section& b) const
    //{
     //   return left < b.left;
    //}
};
bool cmp(const struct section&a, const struct section&b) {
    return a.left < b.left;
}
int main() {
    int n, d,j=1;
    while (scanf("%d%d", &n,&d) && n) {
        int what = 0;
        vector<section>s(n);
        for (int i = 0;i < n;i++) {
            double x, y;
            //cin >> x >> y;
            scanf("%lf%lf",&x,&y);
            if (y > d) { what = 1; continue; }
            if (what)continue;
            double r = sqrt(d * d - y * y);
            s[i].left = x - r;
            s[i].right= x + r;
        }
        if (what) { cout << "Case "<< j++ << ": " << -1 << endl; continue; }
        sort(s.begin(), s.end(),cmp);
        double end = -INT_MAX;
        int radar = 0;
        for (vector<section>::iterator it = s.begin();it != s.end(); it++) {
            if (end < it->left) {//如果上一个区间范围的最右端小于当前区间最左端，则需第二个雷达
                radar++;
                end = it->right;
            }
            else if (end>it->right) {
                end = it->right;
            }
        }
        cout<< "Case " << j++ << ": " << radar << endl;
    }
    return 0;
}