POJ3069左右点贪心 贪心

题目连接

http://poj.org/problem?id=3069

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

题解

从左向右，分两段贪心，记一次数

代码

#include<iostream>  #include<cstdio>  #include<cstring>  #include<string>  #include<cmath>  #include<string> #include<cctype>  #include<algorithm>  #include<vector>#define ll long long#define maxn 1010int num[maxn]={1};int a[maxn];using namespace std;int main(int argc, char const *argv[]){    int n,R;    while(scanf("%d%d",&R,&n)!=EOF){    if(R==-1&&n==-1){        return 0;    }    //memset(a,0,sizeof(a));    for (int i = 0; i < n; ++i)    {        cin>>a[i];    }    int s=0;    int ans=0;    sort(a,a+n);    int i=0;    while(i<n){        s=a[i++];        while(i<n&&a[i]<=s+R)i++;        int p=a[i-1];        ans++;        while(i<n&&a[i]<=p+R)i++;    }    printf("%d\n", ans);}    return 0;}

Get

• 后面while循环的简洁代码