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POJ 1328 Radar Installation(贪心)

2017-08-16 11:44 302 查看
Radar Installation

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 89909 Accepted: 20195
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

Source

Beijing 2002

思路:无解的情况显然很容易判定,只要有一个岛屿到海岸线的距离y>d,那么就无解。对每个岛屿,考虑能监控该岛屿的位置,显然是以该岛屿为圆心,d为半径的圆,与海岸线的两个交点的线段(可能退化)。把所有线段找出来,并按右端排序,若右端相同,左端大的排在右边。贪心策略:从未选择过的线段中选择一个右端点在最左边的线段,在该线段的右端点安装雷达。对进行过贪心选择的区间[a,b]标记覆盖标志,同时对其他的未访问过的区间且左端点坐标小于b的区间也标志覆盖标志,说明在b处安装雷达也可以监控到此岛屿。

代码:

bool cmp(Point a,Point b)
{
if(a.y!=b.y)
return a.y<b.y;
else
return a.x<b.x;
}

void init()
{
for(int i=0;i<n;i++)
{
cin>>point[i].x>>point[i].y;
if(d>=point.y)
{
point[i].x=point[i].x-sqrt(double(d)*d-point[i].y*point[i].y);
point[i].x=point[i].x+sqrt(double(d)*d-point[i].y*point[i].y);
}
else
{
printf("-1\n");
return ;
}
}
}

void solve()
{
sort(point,point+n,cmp);
memset(visit,0,sizeof(visit));
int ans=0;
for(int i=0;i<n;i++)
{
if(!visit[i])
{
visit[i]=1;
ans++;
for(int j=i+1;j<n;j++)
if(!visit[i]&&point[j].x<=point[i].y)//未访问且左端点小于当前右端点
visit[j]=1;
}
}
}
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