POJ1328 Radar Installation 貪心
2015-04-16 10:31
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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
Sample Output
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)。
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
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; }
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