POJ1328 Radar Installation 貪心

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

Source

Beijing 2002


題目大意：在x軸上建設最少的雷達，覆蓋海上的島嶼。
解題思路：考慮每個島嶼若要被覆蓋，則雷達需要建設的範圍。算出每一個島嶼的可建設雷達範圍，進行一次排序（以每個島嶼範圍最佐邊爲排序依據）。因爲每個島嶼必須被覆蓋，所以就某一個島嶼來說，在它被覆蓋在前提下，盡可能的加進去更多的島嶼。所以按照排好的順序枚舉每個島嶼，應用貪心策勵便可計算出答案。 但是要注意一點（在加入新的島嶼時，需要更新這次的右範圍）（博主就是因爲這點wa了幾次，找題解才知道錯因，好悲哀的小菜鳥呀）。

#include <iostream>#include <cstdlib>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;const int maxn = 1e3 + 5;struct node{    double s, t;    bool operator < (const node &a) const {        return s < a.s;    }} nodes[maxn];int main(){    int n, d;    int kase = 1;    while(scanf("%d%d", &n, &d) == 2 && (n || d)) {        int ok = 1;        int x, y;        for(int i = 0; i < n; ++i) {            scanf("%d%d", &x, &y);            if(y > d) {ok = 0; continue;}            nodes[i].s = (double)x - sqrt(d*d - y*y);            nodes[i].t = (double)x + sqrt(d*d - y*y);        }        if(!ok) {printf("Case %d: -1\n", kase++); continue;}        sort(nodes, nodes+n);        double tag = nodes[0].t;        int ans = 1;        for(int i = 0; i < n; ++i) {            if(nodes[i].s <= tag) {tag = min(tag, nodes[i].t); continue;}            else {ans++; tag = nodes[i].t;}        }        printf("Case %d: %d\n", kase++, ans);    }    return 0;}