贪心 + 计算几何 --- Radar Installation

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

【题目来源】

Beijing 2002

http://poj.org/problem?id=1328

【解题思路】

以每个岛的坐标为圆心画圆，会与x轴有2个交点，那么这2个点就是能覆盖该岛的雷达x 坐标区间，问题就转变成对一组区间，找最少数目的点，使得所有区间中都有一点。把包含某区间的区间删掉（如果一个点使得子区间得到满足，那么该区间也将得到满足），这样所有区间的终止位置严格递增。

每次迭代对于第一个区间，选择最右边一个点，因为它可以让较多区间得到满足，如果不选择第一个区间最右一个点（选择前面的点），那么把它换成最右的点之后，以前得到满足的区间，现在仍然得到满足，所以第一个区间的最右一个点为贪婪选择，选择该点之后，将得到满足的区间删掉，进行下一步迭代，直到结束。

ac代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define MAX 1010
struct Node
{
    float x,y;
    float l,r;
    bool vis;
};
Node node[MAX];
int n;
float r;
bool cmp(Node a,Node b)
{
    return a.r<b.r;
}
int main()
{
//    freopen("in.txt","r",stdin);
    int kase=1;
    while(cin>>n>>r)
    {
        if(n==0&&r==0)
            break;
        int i,j;
        int cnt=0;
        if(r<=0)
            cnt=-1;
        for(i=0;i<n;i++)
        {
            scanf("%f%f",&node[i].x,&node[i].y);  //血的教训，这儿用cout绝对超时
            if(node[i].y>r)
                cnt=-1;
            node[i].l=node[i].x-sqrt(r*r-node[i].y*node[i].y);
            node[i].r=node[i].x+sqrt(r*r-node[i].y*node[i].y);
        }
        printf("Case %d: ",kase++);
        if(cnt==-1)
            {cout<<"-1"<<endl;continue;}
        for(i=0;i<n;i++)
            node[i].vis=false;
        sort(node,node+n,cmp);
        bool flag;
        for(i=0;i<n&&cnt>=0;i++)
        {
            if(!node[i].vis)
            for(j=0;j<n;j++)
            {
                if(!node[j].vis)
                {
                    if(node[j].l<=node[i].r)
                    {
                        node[j].vis=true;
                        flag=true;
                    }
                    else break;
                }
            }
            if(flag)
                cnt++;
            flag=0;
        }
        cout<<cnt<<endl;
    }
    return 0;
}