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poj 1328 Radar Installation （贪心）

2024-07-19 10:57:46   228人阅读

Radar Installation
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 52823 Accepted: 11883

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


思路 ：在X轴上肯定有个区域，雷达在这个区域内可以把某个小岛覆盖住；然后把这些区域按左区间排序，然后贪心的求解就行了。

这些仔细考率都能过的，因为一个变量搞成int型竟然错了好久！


#include"stdio.h"
#include"string.h"
#include"math.h"
#include"iostream"
#include"algorithm"
using namespace std;
#define N 1005
struct node
{
    double l,r;
}p[N];
double x[N],y[N];
bool cmp(node a,node b)
{
    return a.l<b.l;
}
int main()
{
    int i,n,d,cnt=1,flag;
    while(scanf("%d%d",&n,&d),n||d)
    {
        flag=0;
        for(i=0;i<n;i++)
        {
            scanf("%lf%lf",&x[i],&y[i]);
            if(flag)
                continue;
            if(d<y[i])
                flag=1;
            double t=sqrt(d*d-y[i]*y[i]);
            p[i].l=x[i]-t;
            p[i].r=x[i]+t;
        }
        printf("Case %d: ",cnt++);
        if(flag)
        {
            printf("-1\n");
            continue;
        }
        sort(p,p+n,cmp);
        int t=1;
        double rr;
        rr=p[0].r;
        for(i=1;i<n;i++)
        {
            if(p[i].l>rr)   //该区间不能被雷达覆盖住
            {
                rr=p[i].r;
                t++;
            }
            if(p[i].r<rr)  //可以被当前雷达作用，
            {
                rr=p[i].r;
            }
        }
        printf("%d\n",t);
    }
    return 0;
}



poj 1328 Radar Installation （贪心）


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