CSU-ACM2017暑期训练7-模拟&&贪心 A

题目：

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 21 2-3 12 11 20 20 0

Sample Output

          Case 1: 2

          Case 2: 1

题意：x轴以上为海洋，x轴及以下为陆地，海洋中有若干个小岛，现想要侦测到他们，给你一种雷达的检测半径，雷达只能放在陆地上，问最少需要多少个雷达能覆盖全所有的小岛。

思路：显然雷达全放在x轴上是离海洋最近的，对于每一个小岛，都有一个在x轴上放置雷达能覆盖到他的区间，求出这个区间的两端，然后贪心求解即可，具体看代码。

代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int INF = 1000000000;const int maxn = 100;int n,d;int ok,no=0;struct Island{    double l,r;}island[1111];bool cmp(Island a,Island b){    return a.l<b.l;}int main(){    while(scanf("%d%d",&n,&d)!=EOF)    {        if(n==0&&d==0)            break;        ok=1;        for(int i=0;i<n;i++)        {            int a,b;            scanf("%d%d",&a,&b);            island[i].l=(double)a-sqrt((double)d*d-b*b);            island[i].r=(double)a+sqrt((double)d*d-b*b);            if(b>d)                ok=0;        }        if(!ok)        {            printf("Case %d: -1\n",++no);            continue;        }        sort(island,island+n,cmp);        int ans=1;        double radar_x=island[0].r;        for(int i=1;i<n;i++)        {            if(radar_x>island[i].r)            {               radar_x=island[i].r;            }            else if(radar_x<island[i].l)            {                radar_x=island[i].r;                ans++;            }        }        printf("Case %d: %d\n",++no,ans);    }    return 0;}