POJ 1328 Radar Installation

雷达覆盖问题，贪心算法。

题目来源：CSUST2012年8月暑假集训组队后第一场个人赛（貌似是8月7日）

原题链接：http://poj.org/problem?id=1328

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 34841		Accepted: 7737

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

                                           

注意：

计算左右交点时的精度问题，全部用double算，tmd 有小数

2013 暑假重做：

/*
2013-7-22晚重做code
题意：直角坐标系上方有 N 个岛屿，有一种雷达，可以覆盖半径为 d 的圆
      现在在 X 轴上设置雷达，问最多修建多少个可以覆盖全部的岛屿
      如果无法覆盖全部的岛屿，则输出 -1
算法：贪心
思路：1.以各个岛屿为圆心，雷达覆盖长度为半径作圆，记录与 X 轴的左右交点
      2.按照各个岛屿与 X 轴的交点排序，先按照左交点从小到大排序，在按照又交点从小到大排序。
      3.排序后从小到大依次遍历每个岛屿，如果有相交的区间，则用同一个雷达覆盖【同时注意缩小相交区间 [left最大的, right最小的]】，
        否则另外增加一个雷达。【就和前面湖大那题一样。。。关于区间的处理】
*/
#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;

const int maxn = 1000+10;
struct Point
{
    double x,y; //坐标
    double left,right; //与 X 轴左右交点
}p[maxn];

bool cmp(Point a, Point b) //按照与 X 轴交点从左到右排序
{
    if(a.left == b.left) return a.right < b.right;
    else return a.left < b.left;
}

int main()
{
    int n,d;
    int test = 0;
    while(scanf("%d %d", &n, &d) != EOF)
    {
        if(n == 0 && d == 0) break;
        int ans = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%lf%lf", &p[i].x, &p[i].y);
            if(p[i].y > d) ans = -1; //一定无法满足覆盖
        }

        if(ans == -1) //输出无法满足的情况
        {
            printf("Case %d: %d\n", ++test, ans);
            continue;
        }

        for(int i = 0; i < n; i++)
        {
            double s = sqrt(d*d-p[i].y*p[i].y);
            p[i].left = p[i].x-s; // 与 X 轴左右交点
            p[i].right = p[i].x+s;
        }
        sort(p,p+n,cmp);

        ans = 1;
        for(int i = 1; i < n; i++)
        {
            if(p[i].left <= p[i-1].right) // 与前面的相交区间
            {
                p[i].left = max(p[i].left,p[i-1].left); // 不断缩小区间
                p[i].right = min(p[i].right,p[i-1].right);
            }
            else ans++;
        }
        printf("Case %d: %d\n", ++test, ans);
    }
    return 0;
}

2012 暑假总结

/******************************************
题意：给你一个直角坐标系;
      在x轴上方给你n个小岛，再给出雷达的覆盖半径d;
      当n和d为0时测试结束;
      剩下n行，依次给出岛屿的坐标（测试数据是int型，注意double转换）;
      求在x轴上最少设置几个雷达，可以覆盖所有的岛屿，如果不能完全覆盖则输出结果为-1;
思路：1、已各个小岛小岛为圆心，以雷达覆盖半径为半径做园，记录与x轴的交点坐标（如果有交点则记录，如果有一个没有交点则表示不能完全覆盖）;
      2、根据各个岛屿与x轴的左右交点排序（先按照左交点从小到大排序，如果坐交点相同，则按照又交点从小到大排序）;
      3、那么经过上面的步骤后，各个交点区肯定会有相加范围，尽量在相加最多的范围设置雷达即可;
      4、从左到右开始查找，设置一个点point为当前雷达覆盖指向的最右点，初始化为第一个圆的右交点;
      5、如果下一个圆的重叠区的左交点大于当前point，雷达数目+1，并且把当前的圆的右交点赋给point;
      6、如果下一个圆的重叠区的右交点小于当前point，说明它已经被覆盖，直接把它的右交点赋给point即可。
注意：由于要用到sqrt，所以要小心double型的转换。
***********************************************/
//Accepted	200K	16MS	C++	928B	2012-09-18 19:43:01
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
struct Node
{
	double left,right;
}node[1010];//圆与x轴的交点
int cmp(Node a,Node b)
{//排序：先按照与x轴的左交点从小到大排，如果左交点相同再按照右交点从小到大排
	return a.left<b.left || (a.left==b.left && a.right<=b.right);
}
int main()
{
	int n;
	int i;
	double d;
	double x,y;
	int sum,ok,c=0;
	double point;//记录当前雷达在x轴上覆盖区域的最右点
	while(scanf("%d%lf",&n,&d)!=EOF && n)
	{
		ok=1;
		if(d<0)//如果雷达覆盖半径<0，则不可能
			ok=0;
		for(i=0;i<n;i++)
		{
			scanf("%lf%lf",&x,&y);
			if(y>d)//排除不可能情况
				ok=0;
			double tt=sqrt(d*d-y*y);//注意double的转换
			node[i].left=1.0*x-tt;//记录左交点
			node[i].right=1.0*x+tt;//记录右交点
		}
		if(ok==0)//如果不可能
		{
			printf("Case %d: -1\n",++c);
		}
		else
		{
			sort(node,node+n,cmp);
			point=node[0].right;
			sum=1;//设置第一个雷达
			for(i=1;i<n;i++)//挨个岛屿查找
			{
				if(node[i].left>point)//如果当前查找的岛屿的最左边的覆盖范围都不在雷达的覆盖范围内
				{
					sum++;//建立新雷达
					point=node[i].right;//以当前的岛屿的右交点赋给新建立的雷达的最右边的覆盖范围
				}
				else if(node[i].right<point)//当前岛屿的右交点都小于前面雷达覆盖位置，则说明已经被覆盖
					point=node[i].right;//更新当前雷达覆盖范围
			}
			printf("Case %d: %d\n",++c,sum);
		}
	}
	return 0;
}