贪心算法

题目描述

1328:Radar Installation
总时间限制: 1000ms 内存限制: 65536kB
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
输入
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
输出
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.
样例输入
3 2
1 2
-3 1
2 1

1 2
0 2

0 0
样例输出
Case 1: 2
Case 2: 1
来源
Beijing 2002

解题分析

将每个小岛被雷达覆盖，需要雷达在海岸线的位置范围表示出来，将所有线段按着起点顺序排序，维护一个被当前雷达覆盖的线段集合，加入一条线段时判断是否与集合中所有线段有交集，如果有，可以加入，如果没有，说明这个线段需要用一个新的雷达点覆盖，更新被当前雷达覆盖的线段集合。
实现的时候可以记录当前集合最小的右端点。

解题代码

#include <cstdio>
#include <algorithm>
#include <cmath>
//#include <vector>
using namespace std;

struct line{
    double l, r;
}s[1005];

bool operator < (const struct line & a, const  struct line & b){
    return a.l < b.l;
}

int n, d;
int main(){
    int Case = 0;
    int ans = 0;
    while(scanf("%d%d", &n, &d) != EOF){
        if(n == 0 && d == 0) break;
        Case++;
        ans = 1;
        int i;
        int flag = 1;
        for(i = 0; i < n; i++){
            int x, y;
            scanf("%d%d", &x, &y);
            if(y > d)
                flag = 0;
            if(flag){
                double a = sqrt(d*d - y*y);
                s[i].l = x - a;
                s[i].r = x + a;
            }
        }
        
        if(!flag) {
            printf("Case %d: %d
", Case, -1);
            continue;
        }
        
        sort(s, s + n);
        double minr = s[0].r;
        
        for(int i = 0; i < n; i++){
            if(s[i].l <= minr){
                minr = s[i].r < minr ? s[i].r : minr;
            }
            else
            {
                ans++;
                minr = s[i].r;
            }
        }
        
        printf("Case %d: %d
", Case, ans);
        
    }
    return 0;
}