Poj 1328 Radar Installation 貪心

題目鏈接：http://poj.org/problem?id=1328

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 52768		Accepted: 11867

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 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

Source

Beijing 2002

源代碼：

#include <iostream>
#include <cmath>
#include <string.h>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <utility>
using namespace std;

const int maxn = 1000 + 5;

int main()
{
        int n, k = 1;
        double d;
        while (scanf("%d%lf", &n, &d) && (n != 0 || d != 0)) {
                vector<pair<double, double> > st;
                bool vis[maxn];
                memset(vis, false, sizeof(vis));
                double x, y, l, r;
                bool flag = false;
                for (int i = 0; i < n; ++i) {
                        scanf("%lf%lf", &x, &y);
                        if (y > d) 
                                flag = true;
                        if (!flag) {
                                l = x - sqrt(d * d - y * y);
                                r = x + sqrt(d * d - y * y);
                                st.push_back(make_pair(r, l));
                        }
                }
                if (flag) {
                        printf("Case %d: -1\n", k++);
                        continue;
                }
                int count = 0;
                sort(st.begin(), st.end());
                memset(vis, false, sizeof(vis));
                for (int i = 0; i < st.size(); ++i) {
                        if (vis[i])
                                continue;
                        ++count;
                        r = st[i].first;
                        vis[i] = true;
                        for (int j = i + 1; j < st.size(); ++j) 
                                if (st[j].second <= r) 
                                        vis[j] = true;
                }
                printf("Case %d: %d\n", k++, count);
        }
}

對於每一個島嶼，在x軸上都有一個雷達可覆蓋其的區間，求最少的雷達，使所有島嶼對應的區間上至少有一個雷達。

將這些區間按照右端點從小到大排序。選擇在右端點建設雷達，進行刷選去除。比如，第一個區間爲[l, r],則剩下區間中左端點小於r的都去掉。

如此類推……