POJ1328 Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 89661		Accepted: 20143

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

 

【题解】

每个点对应一段放置的区间，于是问题转化成了区间点覆盖问题。贪心即可

 

#include <iostream>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <algorithm>
#include <cmath>
#define min(a, b) ((a) < (b) ? (a) : (b))

const int INF = 0x7fffffff;
const int MAXN = 1000 + 10;

inline void read(int &x)
{
    x = 0;char ch = getchar(),c = ch;
    while(ch < '0' || ch > '9')c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
    if(c == '-')x = -x;
}

int n, R, cnt[MAXN], ans, t, ok;
double l[MAXN], r[MAXN], now;

bool cmp(int a, int b)
{
    return l[a] < l[b];
}

int main()
{
    while(scanf("%d%d", &n, &R) && (n + R))
    {
        ans = now = ok = 0;
        register int x,y;
        for(register int i = 1;i <= n;++i)
        {
            cnt[i] = i;
            read(x);read(y);
            if(R < y) ans = -1,ok = 1;
            double tmp = sqrt(R * R - y * y);
            l[i] = x - tmp;
            r[i] = x + tmp;
        }
        if(ok)goto L1;
        std::sort(cnt + 1, cnt + 1 + n, cmp);
        now = -INF;
        for(register int i = 1;i <= n;++ i)
            if(l[cnt[i]] > now) ans ++, now = r[cnt[i]];
            else now = min(now, r[cnt[i]]);
L1:        ;
        if(!ans) ans = -1;
        printf("Case %d: %d
", ++t, ans);
    } 
    return 0;
}