题目链接:http://poj.org/problem?id=1328
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:建立雷达结构体坐标+贪心+线段交集+处理最少的情况.
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn=1010;
struct Island{
double left, right;
}isl[maxn];
bool operator<(const Island &a, const Island &b){
return a.left<b.left; //升序
}
int n, d, x, y;
int solve(){
sort(isl+1, isl+1+n); //按左端点大小升序排列//若当前线段与目前集合中的线段没有交集, 则加入新的雷达
double range = isl[1].right;
int ans=1;
for(int i = 2; i <= n; i++){
if(isl[i].left <= range)
range = min(range, isl[i].right);
else{
range = isl[i].right;
ans++;
}
}
return ans;
}
int main(){
int k=1;
while(1){//n表示具体的点的个数,d表示覆盖距离
cin>>n>>d;
if(n==0&&d==0)
break;
bool flag = 1;
for(int i = 1; i <= n; i++){
scanf("%d%d",&x,&y); //输入具体的坐标
if(y>d)
flag = 0;//此处可以决定输出,即不可能实现的情况
isl[i].left = x*1.0-sqrt(d*d-y*y); //计算每个小岛决定的雷达范围(反向思考)
isl[i].right = x*1.0+sqrt(d*d-y*y); //勾股定理
}
if(flag)
//对于solve()函数的书写十分简单
cout<<"Case "<<k++<<": "<<solve()<<endl;//printf("Case %d: %d\n",k++, solve());
else
cout<<"Case "<<k++<<": -1"<<endl;//printf("Case %d: -1\n",k++); //不可能实现的情况
}
}
/*Sample Input
3 2
1 2
-3 1
2 1
1 2
0 2
0 0
Sample Output
Case 1: 2
Case 2: 1*/