Poj1328Radar Installation (greedy-interval selection)
Question link:
Ah, haha, click me.
Question:
Radar Installation
Time Limit:1000 MS |
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Memory Limit:10000 K |
Total Submissions:52037 |
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Accepted:11682 |
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 abve 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
This question is about interval Selection in greed .. The greedy policy is: the rightmost vertex between constituency.
Train of Thought: first, extract this model, take the island as the center, take the radar distance as the radius, and find the range on land, then the radar is selected
This interval and so on will surely be able to scan the island .. After finding all the intervals, it is converted into the interval selection problem ..
There is also the Standard end in the code to use double, I wa for a long time...
Code:
#include
#include
#include#include
#define INF 0x3f3f3f3fusing namespace std;const int maxn=1000+10;struct Line{ double le,ri;}line[maxn];bool cmp(Line a,Line b){ if(a.ri!=b.ri) return a.ri
b.le;}int main(){ bool ok; int u,v,cas=1; int cnt;double End; int n,d; while(~scanf("%d%d",&n,&d)) { if(n==0&&d==0) return 0; cnt=0; ok=false; for(int i=1;i<=n;i++) { scanf("%d%d",&u,&v); if(v>d) ok=true; else { line[i].le=(double)u-sqrt((double)(d*d-v*v)); line[i].ri=(double)u+sqrt((double)(d*d-v*v)); } } if(ok) printf("Case %d: -1\n",cas++); else { sort(line+1,line+1+n,cmp); cnt=0; End=-INF; for(int i=1;i<=n;i++) { if(line[i].le>End) { cnt++; End=line[i].ri; } } printf("Case %d: %d\n",cas++,cnt); } } return 0;}