Radar Installation - POJ 1328 贪心-优快云博客

本文介绍了一种算法问题，即如何在海岸线上设置雷达站以覆盖海中的所有岛屿，并尽可能减少雷达站的数量。通过确定每个岛屿对应的雷达覆盖区间，采用排序和贪心策略找到了最优解。

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Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 54211		Accepted: 12195

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

题意：在x轴上建立雷达，每个雷达的范围是半径为d的圆，问覆盖所有的岛屿至少需要多少雷达。

思路：对于每个岛屿，我们都可以找到在x轴上的雷达建立范围，要从左往右建的话，就首先找到右端点最小的，之后l小于这个数的都已经覆盖，如果l大于这个数的，就让答案+1，更新右端点。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int n,d;
struct node
{
    double l,r;
    bool flag;
}island[1010];
bool cmp(node a,node b)
{
    return a.r<b.r;
}
bool flag;
int main()
{
    int t=0,i,j,k,x,y,pos,ans;
    double ret,L,R;
    while(~scanf("%d%d",&n,&d) && n>0)
    {
        flag=1;
        t++;
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&x,&y);
            if(y>d)
            {
                flag=0;
                continue;
            }
            island[i].l=x-sqrt((double)(d*d-y*y));
            island[i].r=x+sqrt((double)(d*d-y*y));
            island[i].flag=1;
        }
        if(!flag)
        {
            printf("Case %d: -1\n",t);
            continue;
        }
        sort(island+1,island+1+n,cmp);
        ans=1;
        R=island[1].r;
        for(i=1;i<=n;i++)
        {
            if(island[i].l>R)
            {
                ans++;
                R=island[i].r;
            }
        }
        printf("Case %d: %d\n",t,ans);
    }
}