POJ1328-Radar Installation

描述：

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

　　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

　　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.

代码：

　　每一个海岛可以在产生一个圆心的范围，在该范围内的任意雷达都可以覆盖到该点。要求雷达的最少的数目，采用贪心的思路，就是使选择的圆心能够尽量的属于更多的圆心的区间。如果雷达范围无法到达某个点，则无解。

　　边输入边处理的时候，不能break。就因为这个RE了无数次（╯‵□′）╯︵┴─┴

#include <cmath>
#include <cstdlib>
#include <iostream>
using namespace std;
typedef struct{double left;double right;
}node;
node a[1005];
int cmp(const void *a, const void *b){return (*(node*)a).left >= (*(node*)b).left ? 1 : -1;
}
int main(){int tc=1,n,d,flag,count,x,y;double delta,t;while( scanf("%d%d",&n,&d)!=EOF ){if( n==0 && d==0 ) break;flag=1;if( d<=0 ) flag=0;for( int i=0;i<n;i++ ){scanf("%d%d",&x,&y);if( y<=d ){//岛屿在雷达范围delta=sqrt((double)(d*d-y*y));a[i].left=x-delta;//得到区间a[i].right=x+delta;}else{flag=0;//这里不能写break，因为输入还未结束
            }}if( flag==0 )//无解printf("Case %d: -1\n",tc);else{qsort(a,n,sizeof(node),cmp); t=a[0].right;count=1;for( int i=1;i<n;i++ ){if( a[i].left>t ){//当前区间左界大于相交区间的最右count++;t=a[i].right;//更新相交右界
                }else{if( a[i].right<t )//取相交的区间t=a[i].right;//更新相交区间右界
                }}printf("Case %d: %d\n",tc,count);}tc++;}system("pause");return 0;
}

转载于:https://www.cnblogs.com/lucio_yz/p/4771828.html

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