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
首先计算出每个海岛对应的能够在海岸线上修建雷达站的区间
再按区间的左值排序 题目就转化为了区间选点的贪心问题
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll __int64
#define MAXN 1000
#define INF 0x7ffffff
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
struct Island
{
double l,r;
};
Island p[12000];
int cmp(Island a,Island b)
{
return a.l<b.l;
}
int main()
{
int n,m,i,j,cnt=1;
int sum,ok;
int x,y,r;
while(scanf("%d%d",&n,&r)!=EOF)
{
if(n==0&&r==0) break;
ok= 1;sum=1;
for(i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
if(y>r)ok=0;
if(ok)
{
p[i].l=x-sqrt((double)(r*r-y*y));
p[i].r=x+sqrt((double)(r*r-y*y));
}
}
if(!ok)
{
printf("Case %d: -1\n",cnt++);
continue;
}
sort(p,p+n,cmp);
double mark=p[0].r;
for(i=1;i<n;i++)
{
if(mark<p[i].l)
{
sum++;
mark=p[i].r;
}
if(mark>p[i].r)
mark=p[i].r;
}
printf("Case %d: %d\n",cnt++,sum);
}
return 0;
}
这是一开始的代码 感觉思路也没有错 估计是精度误差太大
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll __int64
#define MAXN 1000
#define INF 0x7ffffff
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
struct Island
{
double x,y;
};
Island p[1200];
int sum;
int cmp(Island a,Island b)
{
if(a.x!=b.x) return a.x<b.x;
return a.y>b.y;
}
int main()
{
int n,m,i,j,cnt=1;
int sum;
double x,y,r;
while(scanf("%d%lf",&n,&r)!=EOF)
{
if(n==0&&r==0) break;
int ok= 1;
sum=1;
for(i=0;i<n;i++)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
if(p[i].y>r) ok=0;
}
if(!ok)
{
printf("-1\n");
continue;
}
sort(p,p+n,cmp);
for(i=0;i<n;i++)
{
if(i==0) x=sqrt(r*r-p[i].y*p[i].y)+p[i].x;
else
{
if(p[i].y*p[i].y+(p[i].x-x)*(p[i].x-x)>r*r)
{
sum++;
x=sqrt(r*r-p[i].y*p[i].y)+p[i].x;
}
}
}
printf("Case %d: %d\n",cnt++,sum);
}
return 0;
}
poj 1328 Radar Installation,布布扣,bubuko.com
原文：http://www.cnblogs.com/sola1994/p/3913756.html