POJ_1328_Radar Installation(greedy)

标签：acm   algorithm   poj   greedy   sort   

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

Output

Sample Input

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

题意：二维坐标系中有n个点，现在要你用最少的圆（圆心在x轴，半径为d）覆盖所有的点。

分析：贪心题。先预处理出每个点对应的圆心坐标区间，意思就是在这个区间内的任意一个点作为圆心坐标都能覆盖该点。然后根据区间右边界升序排序。然后就是区间有无交集的问题了。如果有交集，说明可以用同一个圆覆盖，如果没交集，ans+=1。

题目链接：http://poj.org/problem?id=1328

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;

const int maxn = 1000 + 5;

struct Point{
    double x;
    double y;
}point[maxn];

int n,k=1;
double d,X,Y;
bool judge;

bool cmp(Point a,Point b){
    return a.y<b.y;
}

void input(){
    judge=true;
    for(int i=0;i<n;i++){
        scanf("%lf%lf",&X,&Y);

        if(fabs(Y)>d || d<=0){
            judge=false;
        }
        else{
            point[i].x=X-sqrt(d*d-Y*Y);
            point[i].y=X+sqrt(d*d-Y*Y);
        }
    }
}

void solve(){
    printf("Case %d: ",k++);

    if(!judge || d<=0){
        printf("-1\n");
        return ;
    }
    sort(point,point+n,cmp);
    int ans=1;
    double maxy=point[0].y;
    for(int i=1;i<n;i++){
        if(point[i].x>maxy){
            maxy=point[i].y;
            ans++;
        }
    }
    printf("%d\n",ans);
    return ;
}

int main(){
    while(scanf("%d%lf",&n,&d)!=EOF){
        if(n==0&&d==0) break;
        input();
        solve();
    }return 0;
}

POJ_1328_Radar Installation(greedy)

标签：acm   algorithm   poj   greedy   sort   

原文地址：http://blog.csdn.net/jhgkjhg_ugtdk77/article/details/46128721