【原创】【区间上的贪心 第三篇】Saruman's Army POJ 3069

选点覆盖其他点Saruman’s Army POJ 3069

题目

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.

题意

有n个人，排成一行，给定坐标。我们可以在这些人身上放一个水晶球来监管这些人。如果以某人为圆心，给定的R为半径画圆，圆内有水晶球，就说他被监管到了。求最少放几个水晶球可以监管所有人。

分析

我们把这些点画在数轴上来看，从左边第一个点开始，在谁身上放水晶球来监管他呢？
既然题目要求最小，我们就要把这个水晶球放在尽量离他远的人身上，当然距离不能超过R，这样，放球的人的左边能监管到的人是最多的，就在他身上放球。别忘了他右边也能监管到别人。

代码

#include<cstdio>#include<algorithm>using namespace std;int n,a[1023],R;int main(){    while(scanf("%d%d",&R,&n)!=EOF)    {        if(n==-1&&R==-1) break;        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        sort(a+1,a+1+n);        int now=1,ans=0;        while(now<=n)        {            int x=now++;            while(now<=n&&a[now]-a[x]<=R)now++;            int y=now-1;            while(now<=n&&a[now]-a[y]<=R)now++;            ans++;        }        printf("%d\n",ans);    }}