Radar Installation 贪心

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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[1001];//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-1; i >= 0; i--)
        st.push(coor[i]);
    while(st.size() >= 2)//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[1001],cor_y[1001];
    double add;
    int cnt = 1;
    while(~scanf("%d %d",&t,&r))
    {
        int ok = 1;//判断小岛的坐标是否合法，
        if(t == 0 && r == 0) break;
        for(int i = 0; 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 = 0;
        }
        if(ok == 0)
        {
            printf("Case %d: -1\n", cnt++);
            continue;
        }
        for(int i = 0; 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 0;
}