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.
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輸入
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
輸出
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.
樣例輸入
3 2
1 2
-3 1
2 11 2
0 20 0
樣例輸出
Case 1: 2
Case 2: 1 題意：給定點集S={（xi，yi）i=1.2.3...n}，求用圓心在x軸上，半徑為r的圓覆蓋S所需的圓的最少個數。解題思路：先把給定的xi，yi,r轉化為x軸上的區間，即圓心所在的區間，這樣就轉化為了區間選點問題。先對右端點從小到大排序，右端點相同時，左端點從小到大排序。代碼：
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;struct radar
{double a, b;
}r[1005];bool comp(radar a1, radar a2)
{if(a1.b != a2.b) //對右端點從小到大排序return a1.b < a2.b;return a1.a < a2.a; //右端點相同，左端點從小到大排序
}int main()
{int n, d, i, cas = 0;while(~scanf("%d%d",&n,&d) && (n + d)){double x, y;int flag = 0, m = 0;for(i = 0; i < n; i++){scanf("%lf%lf",&x,&y);if(fabs(y) > d){flag = 1; //有覆蓋不到的點continue;}double diff = sqrt(d * d - y * y);r[m].a = x - diff;r[m++].b = x + diff;}printf("Case %d: ",++cas);if(flag){printf("-1\n");continue;}sort(r, r + m, comp);int cnt = 1, p = 0;for(i = 1; i < m; i++){if(r[i].a <= r[p].b)continue;else{cnt++;p = i;}}printf("%d\n",cnt);}return 0;
}
