poj1328(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

Sample Output

Case 1: 2
Case 2: 1

区间选点问题，贪心策略，对右端点从小到大排序，相同时左端点从大到小排列。优先选取前面的区间右端点，如无法覆盖下一个区间，则选取下一个右端点。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
int dx[] = {0, 1, 0, -1}, dy[] = {-1, 0, 1, 0};
const int maxn = 25010;

struct island
{
    double left, right;
    bool operator < (const island& b) const
    {
        if (right != b.right)
            return right < b.right;
        return left > b.left;
    }
};

int main()
{
    int n, d, kase = 1;
    while (scanf("%d%d", &n, &d) == 2 && (n || d))
    {
        island s[1010];
        int can = 1;
        for (int i = 0; i < n; ++i)
        {
            int x, y;
            scanf("%d%d", &x, &y);
            if (abs(y) > d)
            {
                can = 0;
                continue;
            }
            s[i].left = x - sqrt(1.0*d*d-y*y);
            s[i].right = x + sqrt(1.0*d*d-y*y);
        }
        if (!can)
        {
            printf("Case %d: -1\n", kase++);
            continue;
        }
        sort(s, s+n);
        int ans = 1;
        island tmp = s[0];
        for (int i = 1; i < n; ++i)
            if (s[i].left > tmp.right)
            {
                ++ans;
                tmp = s[i];
            }
        printf("Case %d: %d\n", kase++, ans);
    }
    return 0;
}

算法码上来文章被收录于专栏

公众号「算法码上来」。godweiyang带你学习算法，不管是编程算法，还是深度学习、自然语言处理算法都一网打尽，更有各种计算机新鲜知识和你分享。别急，算法码上来。