poj 1328贪心
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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
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
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
大概题义就是要在x轴上放圆心,使得放最少的圆可以把所有点覆盖,输出圆的个数
#include <iostream> #include <string.h> #include <math.h> #include <stdio.h> #include <algorithm> using namespace std; #define Maxn 10010 struct Node{ double a,b; }A[Maxn]; bool cmp(Node a,Node b){ if(a.a != b.a){ return a.a < b.a; }else{ return a.b < b.b; } } double wenbao(double R,double high){ return sqrt( pow(R,2.0) - pow(high,2.0) ); } int main(){ int N; double R; double a,b; int count = 1; while(cin >> N >> R,N+R){ bool flag = true; for(int i = 0; i < N; i++){ // cin >> a >> b; scanf("%lf%lf",&a,&b);//用scanf,否则会超时 if(fabs(b) > R){ flag = false; }else{ double temp = wenbao(R,b); A[i].a = a - temp; A[i].b = a + temp; } } sort(A,A+N,cmp); Node p = A[0]; int cnt = 1; for(int i = 1; i < N; i++){ if(A[i].a > p.b){ cnt++; p = A[i]; }else if(A[i].b < p.b){//请仔细思考 p = A[i]; } } if(flag){ printf("Case %d: %d\n",count++,cnt); }else{ printf("Case %d: -1\n",count++); } } }
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