POJ3069 贪心法 覆盖问题

Saruman's Army

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6197 Accepted: 3165

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, …, x_n of each troop (where 0 ≤ x_i ≤ 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

遇到的问题和解决思路：

       个人感觉，这道题如果没有先看过书再写的话，肯定不会想到这个是贪心法，而且也不好描述。因为要同时左右都考虑到，所以要既考虑到右边，也考虑到左边。

       具体的解决思路就看代码吧，虽然AC了，不过自己编写还是和书上有点不一样的。（挑战p46）

给出代码：

#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int r,n;int di[1005];void solve(){int res = 0;sort(di ,di + n);for(int i = 0;i < n; i++){int a = di[i];while(a + r >= di[i] && i < n) i++;a = di[--i];//书上这里是 a = di[i-1];while(a + r >= di[i] && i < n)i++;//感觉这一步的i<n绝对不能少，不然超出范围了，di[i]就是随机的一个数字了res++;i--;//书上没有这一步。之所以这一步和书上有区别，是因为书上用的是while（i<n），然而我这里用的是i++}printf("%d\n",res);}int main(){while(scanf("%d%d",&r,&n)!=EOF){if(r == -1 && n == -1)break;memset(di , 0 ,sizeof(di));for(int i = 0;i < n ; i++)scanf("%d",&di[i]);solve();}return 0;}