PKU1328 Radar Installation

/*Radar Installation

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 10128 Accepted: 1990

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*/

 

#include "iostream"

#include "vector"

#include "math.h"

using namespace std;

 

 

class dot

{

public:

double x;

double y;

double l;

double r;

};

 

void solution(vector<dot> k,double d,int num)

{

int result = 1;

dot tmp;

if(d < 0)

{

result = -1;

cout<<"Case "<<num<<": "<<result<<endl;

return;

}

 

for(vector<dot>::iterator iter = k.begin();iter != k.end();iter++)

{

if((*iter).y > d)

{

result = -1;

cout<<"Case "<<num<<": "<<result<<endl;

return;

}

else

{

double s = sqrt(d * d - (*iter).y * (*iter).y);

(*iter).l = (*iter).x - s;

(*iter).r = (*iter).x + s;

}

}

for(vector<dot>::iterator iter = k.begin();iter != k.end();iter++)

{

for(vector<dot>::iterator iters = iter + 1;iters != k.end();iters++)

{

if((*iter).l > (*iters).l)

{

tmp = (*iter);

(*iter) = (*iters);

(*iters) = tmp;

}

else if((*iter).l == (*iters).l)

{

if((*iter).r > (*iters).r)

{

tmp = (*iter);

(*iter) = (*iters);

(*iters) = tmp;

}

}

 

}

}

double left = (*(k.begin())).l,right = (*(k.begin())).r;

for(vector<dot>::const_iterator iter = k.begin() + 1;iter != k.end();iter++)

{

if((*iter).l > right)

{

result++;

right = (*iter).r;

}

else

{

if((*iter).r < right)

{

right = (*iter).r;

}

}

 

 

}

cout<<"Case "<<num<<": "<<result<<endl;

return;

 

}

 

 

int main(void)

{

int n;

double d;

int num = 1;

while(cin>>n>>d)

{

if(n == 0 || d == 0)

{

break;

}

 

else

{

vector<dot> k;

dot m;

for(int i = 0;i < n;i++)

{

cin>>m.x>>m.y;

k.push_back(m);

}

solution(k,d,num++);

}

 

 

 

}

return 0;

}

 

贪心。。首先来看一种通常的情况。。以岛为圆心距离为半径画圆，圆与X轴相交于2个点，这2个点一左一右就产生了一个区间，实际上仔细思考就能明白X轴上处于这个区间内的点就是以这些点任一个为圆心距离d为半径画圆能够把当前这个岛包括进来的雷达的集合。。。那么一个岛的区间和另一个岛的区间若有重合区域。。那么就有雷达能够同时覆盖这2个岛。。如果没有重复区间。。那么就需要2个雷达了。。以此类推。。

以上就是这个题目的中心思想。。。具体做时先判断以岛为圆心d为半径画圆能不能与X轴相交。。如果出现不能相交的立即可以得出结论Case 1: -1（注意本题要判断d为负的情况，d为负时也是直接返回结果-1，不做这个判断这题AC不了）。。。。。然后根据左区间的大小把岛从小到大排序（左区间相同的右区间从小到大）。。这样就可以从左往右来做区间的合并了（当然从右到左也可以）。。。具体区间合并时看我上面的代码。。这题第2个难点就是如何分清楚区间的情况。。最好用纸画一画。。