POJ1328——Radar Installation、POJ1042——Gone Fishing、POJ1456——Supermarket（贪心）

POJ1328——Radar Installation

题目描述

1328 -- Radar Installation

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: 

描述 

假设滑行是一条无限的直线。陆地在海岸的一侧，海洋在另一侧。每个小岛都是位于海边的一个点。而任何雷达装置，位于滑行区，只能覆盖 d 距离，因此如果它们之间的距离最多为 d，则海上的一个岛屿可以被半径装置覆盖。
我们使用笛卡尔坐标系，定义滑行是 x 轴。海边在 x 轴上方，陆地边在 x 轴上方。给定每个岛屿在海中的位置，以及雷达装置的覆盖范围，您的任务是编写一个程序来查找覆盖所有岛屿的最小雷达装置数量。请注意，孤岛的位置由其 x-y 坐标表示。

图 A 雷达装置的输入示例

输入

输入由多个测试用例组成。每种情况的第一行包含两个整数 n （1<=n<=1000） 和 d，其中 n 是海中的岛屿数，d 是雷达装置的覆盖距离。后面跟着 n 行，每行包含两个整数，表示每个岛的位置坐标。然后，空行将大小写分隔开来。
输入以包含一对 0 的行终止

输出

对于每个测试用例，输出一行，由测试用例编号和所需的最小雷达安装数组成。“-1” 安装表示没有针对该情况的解决方案。

样本输入

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

示例输出

Case 1: 2
Case 2: 1

运行代码

代码一语POJ的C++版本不兼容，会出现编译错误，代码二是正确AC的

编译错误

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
const int N = 1005; 
struct Inter {
    double left, right;
};
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, d, c = 1;
    Inter a[N];
    while (cin >> n >> d && (n != 0 || d != 0)) {
        bool p = true;
        for (int i = 0; i < n; ++i) {
            int x, y;
            cin >> x >> y;
            if (y > d || d < 0) {
                p = false;
            }
            else if (d == 0) {
                if (y != 0) p = false;
                else {
                    a[i].left = x;
                    a[i].right = x;
                }
            }
            else {
                double dist = sqrt(d * d - y * y);
                a[i].left = x - dist;
                a[i].right = x + dist;
            }
        }
        if (d < 0) p = false;
        if (!p) {
            cout << "Case " << c<< ": -1\n";
        }
        else {
            sort(a, a + n, [](const Inter& a, const Inter& b) {
                return a.right < b.right;
                });
            int count = 0;
            double last = -INFINITY;
            for (int i = 0; i < n; ++i) {
                if (a[i].left > last) {
                    ++count;
                    last = a[i].right;
                }
            }
            cout << "Case " << c<< ": " << count << "\n";
        }
        ++c;
    }
    return 0;
}

AC 

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;

// 比较函数替代lambda表达式
bool compareIntervals(const pair<double, double>& a, const pair<double, double>& b) {
    return a.second < b.second;
}

int main() {
    int n, d;
    int case_num = 0;
    while (cin >> n >> d && (n != 0 || d != 0)) {
        case_num++;
        vector<pair<int, int> > islands(n);
        for (int i = 0; i < n; i++) {
            cin >> islands[i].first >> islands[i].second;
        }

        bool possible = true;
        int result = 0;

        if (d < 0) {
            possible