POJ-1328（贪心）

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 ddistance, 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 writea 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 InstallationsInputThe 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 isfollowed 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 OutputFor 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

/*
题解：贪心策略
以小岛为圆心画圆，与X轴相交两点
再按照左端点从小到大排序
最后贪心过一遍就OK了

*/
#include <iostream>
#include<math.h>
#include<algorithm>
#include<stdio.h>
using namespace std;
struct pos
{
int x, y;
}nowpos[1005];
int flag = 1;
struct node
{
double left, right;
}nownode[1005];

int d, n;

bool cmp(node ss, node tt)
{
return ss.left < tt.left;
}

void Cal()
{
for (int i = 0; i < n; i++)
{
nownode[i].left = nowpos[i].x - ((double)sqrt(pow(d,2) - pow(nowpos[i].y,2)));
nownode[i].right = nowpos[i].x + ((double)sqrt(pow(d, 2) - pow(nowpos[i].y, 2)));
}
}

int FinaAns()
{
int cnt = 1;
sort(nownode, nownode + n, cmp);//按照左端点从小到大排序
double temp = nownode[0].right;
//贪心算法
for (int i = 1; i < n; i++)
{
//没注意到就Wrong了，重点^_^|||
//后面的右端点可以直接小于之前的右端点
//这样就要更新右端点
if (nownode[i].right < temp)
{
temp = nownode[i].right;
}
else if (temp < nownode[i].left)
{
cnt++;
temp = nownode[i].right;
}
}
return cnt;
}

int main()
{
int tempcount = 0;
while (cin >> n >> d)
{
flag = 1;
if (n == 0 && d == 0)
break;
for (int i = 0; i < n; i++)
{
scanf("%d%d", &nowpos[i].x, &nowpos[i].y);
if (nowpos[i].y > d)
{
flag = 0;
}
}
if (flag == 0)
{
//找了半天的错原来就是这里少了个Case，Q_Q
printf("Case %d: -1\n", ++tempcount);
}
else
{
Cal();//计算
printf("Case %d: %d\n", ++tempcount, FinaAns());//输出答案
}

}

return 0;
}