C-C Radar Installation 解题报告
C-C Radar Installation 解题报告
题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86640#problem/C
题目:
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<tex2html_verbatim_mark> distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d<tex2html_verbatim_mark> .
We use Cartesian coordinate system, defining the coasting is the x<tex2html_verbatim_mark> -axis. The sea side is above x<tex2html_verbatim_mark> -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<tex2html_verbatim_mark> - y<tex2html_verbatim_mark> coordinates.
Input
The input consists of several test cases. The first line of each case contains two integers n<tex2html_verbatim_mark>(1n
1000)<tex2html_verbatim_mark> and d<tex2html_verbatim_mark> , where n<tex2html_verbatim_mark> is the number of islands in the sea and d<tex2html_verbatim_mark> is the distance of coverage of the radar installation. This is followed by n<tex2html_verbatim_mark> 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
题目大意:
在一条海岸线的一侧有若干个岛屿,将海岸线看成是X轴,X轴以上是大海,在海岸上安装雷达使其能覆盖全部岛屿,给出雷达的覆盖半径和岛屿位置,求最少使用多少
雷达才能将所有岛屿全部覆盖。无法覆盖输出-1. 分析:
1.可以换一种想法,求雷达覆盖小岛就是以小岛为圆心,雷达的覆盖半径为半径画圆
2.小岛到海岸线的距离大于半径,不能覆盖所有小岛,输出-1
3.若小于半径,该圆与X轴的左交点为line[i].l=(double)x-sqrt((double)d*d-y*y);右交点为 line[i].r=(double)x+sqrt((double)d*d-y*y);
4.将左交点排序,判断雷达位置,如果i点的左交点在当前雷达的右边,需要安装一个新雷达;如果i点的右交点在当前雷达的左边,则把当前雷达的圆心更新为该点的右交点 代码:
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std; const int maxn=; struct Line //每个岛作半径为d的圆,与x轴所截的线段
{
double l,r;
}line[maxn]; bool cmp(Line a,Line b)//按照线段的左端点从小到大排序
{
return a.l<b.l;
} int main()
{
int n,d;
int i;
int x,y;
bool yes;//确定是不是有解
int m=;
while(scanf("%d%d",&n,&d)!=EOF)
{
yes=true;
int cnt=;
if(n==&&d==) //n,d均为0,程序结束
break;
for(i=;i<n;i++)
{
scanf("%d%d",&x,&y);
if(yes==false)
continue;
if(y>d)
yes=false;
else
{
line[i].l=(double)x-sqrt((double)d*d-y*y);
line[i].r=(double)x+sqrt((double)d*d-y*y);
}
}
if(yes==false) //不能完全覆盖
{
printf("Case %d: -1\n",m++);
continue;
}
sort(line,line+n,cmp);//排序
cnt++;
double now=line[].r;
for(i=;i<n;i++)
{
if(line[i].r<now) //与现在比较,这个很重要
now=line[i].r;
else if(now<line[i].l)
{
now=line[i].r;
cnt++;
}
}
printf("Case %d: %d\n",m++,cnt);
} return ;
}
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