Radar Installation（贪心）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 52444		Accepted: 11792

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：

       给出 n（1 ~ 1000），d，后给出 n 个坐标点，问在 x 轴上选择一些点，这些点能覆盖周围半径 d 以内的地方，问最少在坐标轴上建立几个点，使之能覆盖所有给出的点，如果不能则输出 -1。

  

       思路：

       贪心。找出最少的点数，使之覆盖所有的区间。对于每个给出的点，在半径 d 以内都会在 x 轴上有一个区间范围，找出所有点的区间范围之后。题目变为求解找到最少的点使之所有区间得到覆盖的问题。先对所有区间范围的左坐标由小到大排序一遍，维护最右的端点值，如果下一个区间的左边大于当前右端点值，则需要增加一个点，且维护的最右端点值变为该区间的右端点值。还有一个值得注意的地方就是，如果这个区间的右端点值小于当前维护的右端点值，则需要更新这个端点值。如果在计算区间的时候，遇到 y > d 的则可以判断为 -1 输出。

 

        AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

typedef struct { double a, b; } node;

node no[1005];

double dis (double a, double c) {
    double res = c * c - a * a;
    if(res < 0) return -1;
    return sqrt(res);
}

bool cmp (node a, node b) { return a.a < b.a; }

int main() {

    int n, t = 0;
    double c;

    while (~scanf("%d%lf", &n, &c) && n && c) {
        int temp = 0;
        memset(no, 0, sizeof(no));

        for (int i = 0; i < n; ++i) {
            double x, b;
            scanf("%lf%lf", &x, &b);
            if (b > c) temp = 1;
            else {
                double a = dis(b, c);
                no[i].a = x - a;
                no[i].b = x + a;
            }
        }

        printf("Case %d: ", ++t);
        if (temp) printf("-1\n");
        else {
            sort(no, no + n, cmp);

            int ans = 1;
            double ee = no[0].b;

            for (int i = 1; i < n; ++i) {
                if (no[i].b < ee) ee = no[i].b;
                else if (no[i].a > ee) {
                    ++ans;
                    ee = no[i].b;
                }
            }

            printf("%d\n", ans);

        }
    }

    return 0;
}