POJ 1328 Radar Installation (贪心)
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Radar Installation
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 87083 Accepted: 19500
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
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
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 21 2-3 12 11 20 20 0
Sample Output
Case 1: 2Case 2: 1
题目大意:
x 轴上 分布雷达, 在 轴周围 有许多的点, 这些点是小岛, 问 最少需要多少个雷达,可以把 这些 岛屿覆盖, 如果, 岛屿 距离x轴的距离 超过 雷达半径 则输出-1 ;
刚开始 ,我的思路是 通过 贪心雷达的 位置 来 扫描, 包括点, 一次一次的 移动距离, 后来发现 并不能用,
正解是 通过岛屿,做雷达半径的圆, 与x 轴 交 两个 点 left 和 right 分别比较 岛屿的 left 和right 来 获得 雷达数目, 如果没有交叉 则需要雷达, 有交叉不需要;
#include <iostream>#include <algorithm>#include <stdio.h>#include <cmath>#include <queue>#include <string>typedef long long ll;const int MAXN=1300;using namespace std;struct node{ double l; double r;}a[MAXN],b[MAXN],k;int cmp(node a,node b){ if(a.l==b.l) return a.r>b.r; return a.l<b.l;}int main(){ int n,i,j; double d; int cont=1; while(scanf("%d %lf",&n,&d),n!=0&&d!=0) { int flag=0; for(i=0;i<n;i++)// b[i] l-->x r-->y { cin>>b[i].l>>b[i].r; if(fabs(b[i].r)>d) flag=1; } if(flag||d<0) { printf("Case %d: -1\n",cont++); continue; } sort(b,b+n,cmp); for(i=0;i<n;i++) { a[i].l=b[i].l*1.0-sqrt(d*d-b[i].r*b[i].r); a[i].r=b[i].l*1.0+sqrt(d*d-b[i].r*b[i].r); } k=a[0];int sum=1;// for(i=1;i<n;i++) { if(a[i].l>k.r)//右大于左 无公共区域 { sum++; k=a[i]; } else if(a[i].r<k.r) { k=a[i]; } } printf("Case %d: %d\n",cont++,sum); } return 0;}
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