Poj 1328 Radar Installation 贪心

题目链接：http://poj.org/problem?id=1328

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

Source

源代码：

#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);
        }
}

对于每一个岛屿，在x轴上都有一个雷达可覆盖其的区间，求最少的雷达，使所有岛屿对应的区间上至少有一个雷达。

将这些区间按照右端点从小到大排序。选择在右端点建设雷达，进行刷选去除。比如，第一个区间为[l, r],则剩下区间中左端点小于r的都去掉。

如此类推……

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Poj 1328 Radar Installation 贪心

原文：http://blog.csdn.net/u010826976/article/details/38709251