| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 50843 | Accepted: 11415 |
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
思路：该题题意是为了求出能够覆盖所有岛屿的最小雷达数目，每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达，则可以覆盖该小岛，区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样，问题即转化为已知一定数量的区间，求最小数量的点，使得每个区间内斗至少存在一个点。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。
以左区间为准：
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
const int M = 1000 + 5;
double x, y;
double dis; //圆的半径
double temp;
int ans; //用于统计结果
int n; //点的个数
int flag; //用于判断
struct node
{
double left; //左区间
double right; //右区间
} island[M];
bool cmp(node a, node b) //按左区间从小到大排序
{
return a.left<b.left;
}
int main()
{
int cas=0;
while(scanf("%d%lf", &n, &dis) && n && dis)
{
flag=0;
ans=0;
cas++;
for(int i=0; i<n; i++)
{
scanf("%lf%lf", &x, &y);
if(y>dis||y<0) //不符合的点
flag=1;
island[i].right = x+sqrt(dis*dis-y*y);
island[i].left = x-sqrt(dis*dis-y*y);
}
if(flag==1)
printf("Case %d: -1\n",cas);
else
{
sort(island, island+n, cmp); //sort排序
temp = island[0].right;
ans=1;
for(int i=1; i<n; i++)
{
if(island[i].right<temp) //说明这个区间被上个区间所包含，只要小区间满足的点，大区间肯定也满足
temp=island[i].right;
else if(island[i].left>temp) //说明这两个区间没有重合部分，也就不能用一个圆表示，则结果加一
{
ans++;
temp=island[i].right;
}
}
printf("Case %d: %d\n", cas, ans);
}
}
return 0;
}
一右区间为准：（这是我同学写的右区间）
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
struct node
{
double a,b;
}s[1005],l[1005];
int cmp(node x,node y)
{
return x.b<y.b;
}
int main()
{
int n,m,k=0;
while(cin>>n>>m)
{
if(n==0&&m==0) break;
int i,j,p=0;
for(i=0;i<n;i++)
{
scanf("%lf%lf",&s[i].a,&s[i].b);
if(s[i].b<0||s[i].b>m)
p=1;
l[i].a=s[i].a-sqrt(m*m-s[i].b*s[i].b);
l[i].b=s[i].a+sqrt(m*m-s[i].b*s[i].b);
}
sort(l,l+n,cmp);
int sum=1; double r=l[0].b;
for(i=1;i<n;i++)
{
if(l[i].a>r)
{
r=l[i].b;
sum++;
}
}
printf("Case %d: ",++k);
if(p)
cout<<-1<<endl;
else
cout<<sum<<endl;
}
return 0;
}
POJ 1328：Radar Installation