POJ1328 Radar Installation 【贪心·区间选点】

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 54593		Accepted: 12292

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

Source

Beijing 2002

区间选点：按左端排序，flag控制右端，若下一项的左端比flag小则看其右端，若右端也比flag小则更新flag，若下一项的左端比flag大，更新flag为这一项的左端并++ans。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>

#define maxn 1010
using namespace std;

struct Node {
double u, v;
friend bool operator<(const Node& a, const Node& b) {
return a.u < b.u;
}
} E[maxn];
int N, D;

int main() {
int i, ok, id, ans, cas = 1;
double x, y, d, flag;
while(scanf("%d%d", &N, &D), N) {
printf("Case %d: ", cas++);
ok = 1; id = 0;
for(i = 0; i < N; ++i) {
scanf("%lf%lf", &x, &y);
if(y > D) ok = 0;
if(!ok) continue;
d = sqrt(D * D - y * y);
E[id].u = x - d;
E[id++].v = x + d;
}

if(!ok) {
printf("-1\n");
continue;
}

sort(E, E + id);

flag = E[0].v; ans = 1;
for(i = 1; i < N; ++i) {
if(E[i].u <= flag) {
if(E[i].v <= flag) flag = E[i].v;
continue;
}
++ans; flag = E[i].v;
}

printf("%d\n", ans);
}
return 0;
}