poj 1328 Radar Installation 【贪心】

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 64022 Accepted: 14402

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

无语死了，以后除了特殊情况不会在杭电DIY做poj的题目了。经常卡死，代码提交还是空代码。 转战zoj。。。

贪心：求出能够覆盖当前点的 区间[Ai，Bi]，按区间右边界 Bi 从小到大排序后， 判断区间是否重叠。

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define INF 100000000#define DD doubleusing namespace std;struct rec{    DD start, end;};rec num[1010];bool cmp(rec a, rec b){    return a.end < b.end;}int main(){    int N, k = 1;    DD d;    bool flag;    while(scanf("%d%lf", &N, &d), N||d)    {        DD x, y;        flag = true;        for(int i = 0; i < N; i++)        {            scanf("%lf%lf", &x, &y);            if(!flag) continue;            if(y > d)            {                flag = false;                continue;            }            num[i].start = x - sqrt(d * d - y * y);            num[i].end = x + sqrt(d * d - y * y);        }        printf("Case %d: ", k++);        if(!flag)        {            printf("-1\n");            continue;        }        sort(num, num+N, cmp);        int ans = 0;        DD temp = -INF;        for(int i = 0; i < N; i++)        {            if(temp < num[i].start)                ans++, temp = num[i].end;        }        printf("%d\n", ans);    }    return 0;}