贪心算法三个经典例题

贪心算法的三个经典例题

A- Saruman’s Army

题目描述：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.

代码示例：

#include 
#include 

using namespace std;

int solve(int a[], int n,int r)
{
    int ans = 0,i = 0;
    sort(a,a+n);       // 将数组升序
    while(i < n)
    {
        int s = a[i++];     // 一直向右前进直到距s的距离大于r的距离的点
        while(i < n&&a[i] <= s + r) i++;
        int k = a[i-1];     // 一直向右前进直到距k的距离大于r的距离的点
        while(i < n&&a[i] <= k + r) i++;
        ans++;
    }
    return ans;
}
int main(void)
{
    int n,m;
    while(cin>> n >> m,~n&&~m)
    {
        int a[m];
        for(int i = 0; i != m; ++i)
        {
            cin >> a[i];
        }
        cout << solve(a,m,n) << endl;
    }
    return 0;
}

B- Fence Repair
题目描述：

代码示例：

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
int main()
{
#ifdef ONLINE_JUDGE  
#else  
    freopen("D:\\in.txt", "r", stdin);
    freopen("D:\\out.txt", "w", stdout);
#endif // ONLINE_JUDEG
    priority_queue<int,vector<int>,greater<int> > pq;
    int n(0);
    cin >> n;//忽略第一行
    while (cin >> n)
    {
        pq.push(n);
    }
    long long sum(0);
    int l1(0), l2(0);
    while (pq.size() > 1)
    {
        l1= pq.top();
        pq.pop();
        l2= pq.top();
        pq.pop();
        sum = sum+l1 + l2;
        pq.push(l1 + l2);
    }
    cout << sum;
    return 0;
}

C–Safe Or Unsafe
题目描述：

代码如下：

#include
#include
#include
using namespace std;
//char str[10010];
//int num[30];
int cmp(int a,int b)
{
	return a>b;
}
int main()
{char str[10010];
    int num[30];
	int i,j;
	int n,m;
	int sum;
	scanf("%d",&n);
	while(n--)
	{
		memset(num,0,sizeof(num));
		scanf("%d",&m);
		scanf("%s",str);
		for(i=0;str[i]!='\0';i++)
		{
			num[str[i]-'a']++;
		}
		sort(num,num+26,cmp);
		for(i=0;i<26;i++)
		{
			if(num[i]==0)
			{
				break;
			}
		}
		sum=0;
		if(i==1)	// 单一字符
		{
			sum=num[0];
		}
		else
		{
			for(j=i-1;j>=1;j--)
			{
				sort(num,num+j+1,cmp);	// 每次都排序取最小 
				sum+=num[j]+num[j-1];
				num[j-1]+=num[j];
			}
		}
		//printf("%d\n",sum);
		if(sum<=m)
		{
			printf("yes\n");
		}
		else
		{
			printf("no\n");
		}
	}
	return 0;
}