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  • POJ 3069 Saruman's Army(贪心)

     Saruman's Army
    Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
    Appoint description: 

    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




    尽量使每个球覆盖的位置最大。
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <string>
     5 #include <algorithm>
     6 #include <cctype>
     7 #include <cmath>
     8 #include <queue>
     9 #include <map>
    10 #include <cstdlib>
    11 using    namespace    std;
    12 
    13 int    main(void)
    14 {
    15     int    r,n;
    16     int    s[1005];
    17     int    ans,left,i,mid;
    18 
    19     while(scanf("%d%d",&r,&n) && (r != -1 && n != -1))
    20     {
    21         for(i = 0;i < n;i ++)
    22             scanf("%d",&s[i]);
    23         sort(s,s + n);
    24 
    25         ans = i = 0;
    26         while(i < n)
    27         {
    28             left = s[i ++];
    29             for(;i < n && s[i] - left <= r;i ++);
    30             left = s[i - 1];
    31             ans ++;
    32             for(;i < n && s[i] - left <= r;i ++);
    33         }
    34         printf("%d
    ",ans);
    35     }
    36 
    37     return    0;
    38 }
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  • 原文地址:https://www.cnblogs.com/xz816111/p/4463930.html
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