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poj 1328 Radar Installation
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 2
1 2
-3 1
2 1
1 2
0 2
0 0
Sample Output
Case 1: 2
Case 2: 1
首先计算出每个海岛对应的能够在海岸线上修建雷达站的区间
再按区间的左值排序 题目就转化为了区间选点的贪心问题
#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#include<queue>#include<stack>#define mem(a,b) memset(a,b,sizeof(a))#define ll __int64#define MAXN 1000#define INF 0x7ffffff#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;struct Island{ double l,r;};Island p[12000];int cmp(Island a,Island b){ return a.l<b.l;}int main(){ int n,m,i,j,cnt=1; int sum,ok; int x,y,r; while(scanf("%d%d",&n,&r)!=EOF) { if(n==0&&r==0) break; ok= 1;sum=1; for(i=0;i<n;i++) { scanf("%d%d",&x,&y); if(y>r)ok=0; if(ok) { p[i].l=x-sqrt((double)(r*r-y*y)); p[i].r=x+sqrt((double)(r*r-y*y)); } } if(!ok) { printf("Case %d: -1\n",cnt++); continue; } sort(p,p+n,cmp); double mark=p[0].r; for(i=1;i<n;i++) { if(mark<p[i].l) { sum++; mark=p[i].r; } if(mark>p[i].r) mark=p[i].r; } printf("Case %d: %d\n",cnt++,sum); } return 0;}
这是一开始的代码 感觉思路也没有错 估计是精度误差太大
#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#include<queue>#include<stack>#define mem(a,b) memset(a,b,sizeof(a))#define ll __int64#define MAXN 1000#define INF 0x7ffffff#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;struct Island{ double x,y;};Island p[1200];int sum;int cmp(Island a,Island b){ if(a.x!=b.x) return a.x<b.x; return a.y>b.y;}int main(){ int n,m,i,j,cnt=1; int sum; double x,y,r; while(scanf("%d%lf",&n,&r)!=EOF) { if(n==0&&r==0) break; int ok= 1; sum=1; for(i=0;i<n;i++) { scanf("%lf%lf",&p[i].x,&p[i].y); if(p[i].y>r) ok=0; } if(!ok) { printf("-1\n"); continue; } sort(p,p+n,cmp); for(i=0;i<n;i++) { if(i==0) x=sqrt(r*r-p[i].y*p[i].y)+p[i].x; else { if(p[i].y*p[i].y+(p[i].x-x)*(p[i].x-x)>r*r) { sum++; x=sqrt(r*r-p[i].y*p[i].y)+p[i].x; } } } printf("Case %d: %d\n",cnt++,sum); } return 0;}