poj1328求大神 题意如下

Question

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

题意：假设海岸线是一条无限延伸的直线。陆地在海岸线的一侧，而海洋在另一侧。每一个小的岛屿是海洋上的一个点。雷达坐落于海岸线上，只能覆盖d距离，所以如果小岛能够被覆盖到的话，它们之间的距离最多为d。

题目要求计算出能够覆盖给出的所有岛屿的最少雷达数目。

 #include <algorithm>
#include <vector>
#include <limits>
#include <string>

using namespace std;
double d;
int n;
int t=0;
bool flag = false;
vector <int> ans;
struct data
{
    public:
    double start;
    double ends;
};

bool compare(data a,data b)
{
    return a.ends<b.ends;
}

int main()
{
    while(cin>>n>>d&&(n!=0&&d!=0))
    {
        vector <data> ils (n);
        int x,y,i;
        i=0;
        while(i!=n&&cin>>x>>y)
        {
            if(y>d)
            {
                flag = true;
            }
            data item;
            item.start = x - sqrt(d*d-y*y);
            item.ends = x + sqrt(d*d-y*y);
            ils[i] = item;
            ++i;
        }
        sort(ils.begin(),ils.end(),compare);
    i=0;
    int num = 0;
    double range;
    while(i<n)
    {
        range = ils[i].ends;
        while(++i<n)
        {
            if(ils[i].start>range)
            break;
        }
        ++num;
    }
    if(flag)
    {
        ans.push_back(-1);
    }
    else
    {
        ans.push_back(num);
    }
    n=d=i=num=0;
    ils.clear();
    flag = false;
    }
    for(t=0;t!=ans.size();t++)
    {
        cout<<"case "<<t+1<<": "<<ans[t]<<endl;
    }

    return 0;
}

Answer 1

你输出的”case“应该改为“Case”，哈哈哈，给跪了

 cout<<"Case "<<t+1<<": "<<ans[t]<<endl;