POJ_1328_Radar Installation_贪心

每次打代码打得特别惨我就想找个人类聊聊天。

题意：

x轴上方为海，其中某些整点位置存在一共n个岛屿，在x轴（海岸）上安排一些灯塔，每个灯塔的扫描半径都为d，求最少需要多少个灯塔。

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.

一开始贪错了，按先x正向后y反向的顺序排，再找，结果WA了几遍，上厕所的时候想明白了反例。

用勾股定理可以求得每个岛在x轴上有一段范围灯塔在其中就可以扫到，因此岛的坐标其实就是这么一个区间，然后就变成了最少点覆盖区间的经典贪心问题了。

代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<algorithm>using namespace std;#define mxn 1010bool dcmp(double a,double b){return fabs(a-b)<0.0001;}double d;struct point{double x,y;bool operator < (const point& in)const{return x+sqrt(d*d-y*y)<in.x+sqrt(d*d-in.y*in.y);}point(){}point(double _x,double _y){x=_x,y=_y;}}p[mxn];bool flag[mxn];int n;int main(){int cs=0;while(scanf("%d%lf",&n,&d)!=EOF){if(!n&&dcmp(d,0))break;memset(flag,0,sizeof(flag));int ans=0;for(int i=0;i<n;++i){scanf("%lf%lf",&p[i].x,&p[i].y);if(p[i].y>d)ans=-1;}if(ans==-1){printf("Case %d: -1\n",++cs);continue;}sort(p,p+n);for(int i=0;i<n;++i){if(flag[i])continue;flag[i] = true;double now=p[i].x+sqrt(d*d-p[i].y*p[i].y);for(int j=i+1;j<n;++j){double cur=p[j].x-sqrt(d*d-p[j].y*p[j].y);if(cur<now||dcmp(cur,now)){flag[j]=true;continue;}break;}++ans;}printf("Case %d: %d\n",++cs,ans);}return 0;}