POJ 3069 Saruman's Army 貪心

標籤：des   style   class   blog   code   http   

Saruman‘s Army Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3612   Accepted: 1846 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 0 310 20 2010 770 30 1 7 15 20 50-1 -1 Sample Output 24 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 Stanford Local 2006

題解

題目抽象出來就是用一個半徑為R的圓去覆蓋一條直線上的點。每個園內必須有至少一個被標記的點。問最少需要標記多少個點。

貪心去處理就行了，給從左邊開始，在園內的最右邊的點加上標記即可。

程式碼範例

/**============================================================================#       COPYRIGHT NOTICE#       Copyright (c) 2014 All rights reserved#       ----Stay Hungry Stay Foolish----##       @author       :Shen#       @name         :POJ 3069#       @file         :G:\My Source Code\【ACM】訓練\0624 - 基礎\poj3069.cpp#       @date         :2014-06-24 14:33#       @algorithm    :Greedy============================================================================**///#pragma GCC optimize ("O2")//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cmath>#include <cstdio>#include <string>#include <cstring>#include <iomanip>#include <iostream>#include <algorithm>using namespace std;template<class T>inline bool updateMin(T& a, T b){ return a > b ? a = b, 1: 0; }template<class T>inline bool updateMax(T& a, T b){ return a < b ? a = b, 1: 0; }typedef long long int64;int r, n;int x[1005];void solve(){    for (int i = 0; i < n; i++)        scanf("%d", &x[i]);    sort(x, x + n);    int i = 0, ans = 0;    while (i < n)    {        int s = x[i++];        while (i < n && x[i] <= s + r) i++;        int p = x[i - 1];        while (i < n && x[i] <= p + r) i++;        ans++;    }    printf("%d\n", ans);}int main(){    while (scanf("%d%d", &r, &n))    {        if (r == -1 && n == -1) break;        else solve();    }    return 0;}