• Saruman`s Army(POJ 3069)


    Saruman's Army
    Time Limit: 1000MS        Memory Limit: 65536K
    Total Submissions: 13974        Accepted: 7011
    Description

    Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. To keep track of his forces, Saruman distributes seeing stones, known as palantirs, among the troops. Each palantir has a maximum effective range of R units, and must be carried by some troop in the army (i.e., palantirs are not allowed to “free float” in mid-air). Help Saruman take control of Middle Earth by determining the minimum number of palantirs needed for Saruman to ensure that each of his minions is within R units of some palantir.

    Input

    The input test file will contain multiple cases. Each test case begins with a single line containing an integer R, the maximum effective range of all palantirs (where 0 ≤ R ≤ 1000), and an integer n, the number of troops in Saruman’s army (where 1 ≤ n ≤ 1000). The next line contains n integers, indicating the positions x1, …, xn of each troop (where 0 ≤ xi ≤ 1000). The end-of-file is marked by a test case with R = n = −1.

    Output

    For each test case, print a single integer indicating the minimum number of palantirs needed.

    Sample Input

    0 3
    10 20 20
    10 7
    70 30 1 7 15 20 50
    -1 -1
    Sample Output

    2
    4

    将点从小到大排序,从左边开始选点,排除覆盖的点,再从新开始选点。

    #include<iostream>
    #include<algorithm>
    using namespace std;
    int main()
    {
        int r,n,d=0;
        int data[1010];
        while(cin>>r>>n&&(r!=-1||n!=-1))
        {
            d++;
            for(int i=0;i<n;i++)
                cin>>data[i];
            sort(data,data+n);
            int i=0;
            int a=data[0];
            int b=data[n-1];
            int total=0;
            while(i<n&&a<=b)
            {
                while(i<n&&data[i]<=a+r)i++;
                a=data[i-1];
                while(i<n&&data[i]<=a+r)i++;
                total++;
                if(i<n)a=data[i];
            }
            cout<<total<<endl;
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/ke-yi-/p/10175834.html
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