zoukankan      html  css  js  c++  java
  • poj 3069 Saruman's Army

    Saruman's Army
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 5760   Accepted: 2949

    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

    Hint

    In the first test case, Saruman may place a palantir at positions 10 and 20. Here, note that a single palantir with range 0 can cover both of the troops at position 20.

    In the second test case, Saruman can place palantirs at position 7 (covering troops at 1, 7, and 15), position 20 (covering positions 20 and 30), position 50, and position 70. Here, note that palantirs must be distributed among troops and are not allowed to “free float.” Thus, Saruman cannot place a palantir at position 60 to cover the troops at positions 50 and 70.

    Source

     1 #include<cstdio>
     2 #include<cstring>
     3 #include<iostream>
     4 #include<stack>
     5 #include<set>
     6 #include<map>
     7 #include<queue>
     8 #include<algorithm>
     9 using namespace std;
    10 int army[1005];
    11 int main(){
    12     //freopen("D:\INPUT.txt","r",stdin);
    13     int n,r;
    14     while(scanf("%d %d",&r,&n)!=EOF){
    15         int i,j,k;
    16         if(n==-1&&r==-1){
    17             break;
    18         }
    19         for(i=0;i<n;i++){
    20             scanf("%d",&army[i]);
    21         }
    22         sort(army,army+n);
    23         int s,m,e,t=0;
    24         for(s=0;s<n;t++){
    25             m=s;
    26             while(m<n&&army[m]-army[s]<=r){
    27                 m++;
    28             }
    29             e=m;
    30             m--;
    31             while(e<n&&army[e]-army[m]<=r){
    32                 e++;
    33             }
    34             s=e;
    35         }
    36         printf("%d
    ",t);
    37     }
    38     return 0;
    39 }
  • 相关阅读:
    C# 控制台应用程序输出颜色字体[更正版]
    ORM for Net主流框架汇总与效率测试
    php 去掉字符串的最后一个字符
    bzoj1185 [HNOI2007]最小矩形覆盖 旋转卡壳求凸包
    bzoj [Noi2008] 1061 志愿者招募 单纯形
    bzoj1009 [HNOI2008] GT考试 矩阵乘法+dp+kmp
    扩展欧几里得(ex_gcd),中国剩余定理(CRT)讲解 有代码
    BZOJ 2103/3302/2447 消防站 树的重心【DFS】【TreeDP】
    hihocoder 1449 后缀自动机三·重复旋律6
    hihocoder 后缀自动机二·重复旋律5
  • 原文地址:https://www.cnblogs.com/Deribs4/p/4854910.html
Copyright © 2011-2022 走看看