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Radar Installation
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 42461 | Accepted: 9409 |
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
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<stack>
#include<math.h>
using namespace std; struct node
{
double start;
double end;
}coor[];//记录每个区间的端点
int cmp(const struct node a,const struct node b)
{
return a.start < b.start;
}
int t,r;
stack <node> st;//用栈存每个区间， int cal(int ans)
{
while(!st.empty())
st.pop();
for(int i = t-; i >= ; i--)
st.push(coor[i]);
while(st.size() >= )//当栈中至少存在两个区间时
{
struct node tmp1 = st.top();
st.pop();
struct node tmp2 = st.top();
if(tmp1.end >= tmp2.start)//当取出的两个区间有公共部分时
{
st.pop();//tmp2出栈
struct node tmp;
tmp.start = max(tmp1.start, tmp2.start);//注意取公共部分时，起始点取较大者
tmp.end = min(tmp1.end, tmp2.end);//终点取较小者
st.push(tmp);//将公共部分入栈
ans--;//每两个区间交一次，雷达个数减一次
}
}
return ans;
}
int main()
{ int cor_x[],cor_y[];
double add;
int cnt = ;
while(~scanf("%d %d",&t,&r))
{
int ok = ;//判断小岛的坐标是否合法，
if(t == && r == ) break;
for(int i = ; i < t; i++)
{
scanf("%d %d",&cor_x[i],&cor_y[i]);
if(cor_y[i] > r)//若小岛纵坐标大于半径则不合法
ok = ;
}
if(ok == )
{
printf("Case %d: -1\n", cnt++);
continue;
}
for(int i = ; i < t; i++)
{
//以每个小岛为圆心，r为半径画圆，coor[]存该圆与x轴相交的区间
add = sqrt(r*r-cor_y[i]*cor_y[i]);
coor[i].start = cor_x[i] - add;
coor[i].end = cor_x[i] + add;
}
sort(coor,coor+t,cmp);//对这些区间按起始点从小到大排序
int ans = t;
ans = cal(ans);
printf("Case %d: %d\n",cnt++,ans);
}
return ;
}