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本文探讨了UVA1193竞赛题目，即如何使用最少数量的雷达安装来覆盖海上的所有岛屿。文章详细介绍了问题背景，输入输出格式，并提供了一段C++代码实现解决方案，通过计算每个岛屿的覆盖范围并使用贪心算法找到最优解。

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UVA 1193

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 1: A Sample Input of Radar Installation
Input Theinputconsistsofseveraltestcases. Thefirstlineofeachcasecontainstwointegersn (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

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
struct point
{
    int x,y;
    double xl,xr;
};
bool comp(point a,point b)
{
    return a.x<b.x;
}
int n,d;
point pt[1005];
int solve(){
    for(int i=0;i<n;i++)
        if(pt[i].y>d)return -1;  //雷达无法侦查到
    sort(pt,pt+n,comp);
    double r=pt[0].xr;
    int ans=1;
    for(int i=1;i<n;i++){
        if(pt[i].xl>r){      //当区间不重叠
            r=pt[i].xr;
            ans++;
        }
        else if(pt[i].xl<r&&pt[i].xr<r)  //区间重叠可以忽略
            r=pt[i].xr;
    }
    return ans;
}
int main()
{
    int cas=1;
    while(scanf("%d%d",&n,&d)==2&&n&&d){
        for(int i=0;i<n;i++){
            scanf("%d %d",&pt[i].x,&pt[i].y);
            double c=sqrt((double)d*(double)d-(double)pt[i].y*(double)pt[i].y);
            pt[i].xl=(double)pt[i].x-c;  //将岛变为侦查区间
            pt[i].xr=(double)pt[i].x+c;
        }
        printf("Case %d: %d\n",cas++,solve());
    }
}