Radar Installation
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打开题目链接:http://poj.org/problem?id=1328
这也算是贪心吗,要是这样的话,贪心看起来好像也很简单啊,行了,贪心就先练到这吧
本题大意:
这题是说,在一条直线两侧分别是海跟陆地,也就是说这条线是海岸线。在海里有若干个用x,y坐标标示位置的岛屿,岸上安排雷达站,每个雷达站有覆盖范围。要求求出最少需要多少个雷达站覆盖所有岛屿
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int M = 1001;struct xy //岛屿坐标{ double l; double r;}p[M];double min(double a,double b){ return a < b ? a :b;}int d,n;bool cmp(struct xy a,struct xy b ){ return a.l < b.l;}int main(){ double a,b; int icase = 0; while(scanf( "%d%d",&n ,&d)&& !(n==0 && d==0) )//n为岛的个数,d为能探测到的半径 { memset( p, 0, sizeof( p ) ); icase ++; int sum = 1; bool flag = false; for(int i = 1; i <= n;i++ ) { cin>>a>>b; if(b>d) ///当有岛屿离海岸的距离大于雷达覆盖半径,则输出不可能。 flag = true; else { p[i].l = a-sqrt(d*d - b*b); p[i].r = a+sqrt(d*d - b*b); } } if(flag) //大于半径 { printf("Case %d: -1\n",icase); continue; } else { sort(p+1,p+n+1,cmp); double s = p[1].r; for(int i = 2; i <= n; i++) ///从左向右 { if(p[i].l > s) { sum++; s = p[i].r; } else s = min(s,p[i].r); } } printf("Case %d: %d\n",icase,sum); } return 0;
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- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
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