poj1328贪心

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 76566 Accepted: 17139

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

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分析：

１．以d为斜边作直角三角形，可知每个坐标在x轴上形成一个闭区间，在此区间内的雷达满足要求；

２．因此题目转化为求区间的集合数，相交的为一个集合，求最小集合数；

３．贪心规则：由左向右摆放雷达，因此从考虑最靠左的区间右端点开始，向右推进求集合数。

<span style="font-size:14px;">#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <cmath>using namespace std;int n;int d;pair<double , double>group[1000];pair<signed int , signed int>coordinate[1000];int times = 0;int main(int argc, char const* argv[]){    //freopen("input.txt","r",stdin);    while (scanf("%d%d",&n,&d)){        if((n == 0) && (d == 0)) break;        times++;        int ans = 1;        bool out = false;        if(d < 0){            out = true;        }        for (int i = 0; i < n; i += 1){            scanf("%d%d",&coordinate[i].first,&coordinate[i].second);            if (d >= coordinate[i].second){                double dd = sqrt(d*d-(coordinate[i].second)*(coordinate[i].second));                group[i].first = coordinate[i].first+dd;//group.right                group[i].second = coordinate[i].first-dd;//group.left            }            else{                out = true;            }        }        if(out){            printf("Case %d: -1\n",times);            continue;        }        sort(group,group+n);//depended on right        int i= 0;        for (int j = i+1; j < n; j += 1){            if (group[j].second > group[i].first){                ans++;                i = j;            }        }        printf("Case %d: %d\n",times,ans);    }    return 0;}</span>