poj3069 Saruman's Army 贪心策略【挑战程序设计竞赛】

题目链接：http://poj.org/problem?id=3069

Saruman’s Army

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 18781 Accepted: 9217
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

Stanford Local 2006

解题思路

1.从左往右 选择一个点i，找区间[i,R] 之内离点i最远的那个点t，将此点放入vector
2.将点t作为标记点，然后找[t,R]之外的第一个点e
-----2.1 若点e在没有越界，则i = e，继续上述过程
-----2.2 若点e已经越界，则说明已经计算完毕了

AC代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const ll maxN = 1e6+10; 

int R,N;
int pos[2000];

vector<int> fun(){ //将选的点放入vector并返回 区间[i,t,e)  i起始点，t标记点，e结束点(下一个区间的起始点)
	vector<int> res;
	int i = 0;//起始位置 
	while(i<N){
		int t = i;
		while(t<N && pos[t]-pos[i]<=R){//找距i点R之外的第一个点t 
			t++;
		}
		res.push_back(pos[t-1]); //所以点t-1是点i R之内最远的那个点，应该将此点作为标记点 
		int e = t - 1;//再将标记点作为起始点，去找下一个i的起始位置 
		while(e<N && pos[e]-pos[t-1]<=R){//找距离标记点第一个R之外的点 
			e++;
		}
		i = e;//更新i，如果此时已经到了N，就结束了 
	}
	return res;
}

int main(){
	while(cin>>R>>N && !(R==-1 && N==-1)){
		for(int i = 0;i<N;i++) scanf("%d",&pos[i]);
		sort(pos,pos+N);
		vector<int> res = fun();
		cout<<res.size()<<endl;//然后vector的个数，即为答案 
	}
	
	return 0;
}