Hackerrank ：Candies

最新推荐文章于 2022-12-25 16:16:14 发布

原创最新推荐文章于 2022-12-25 16:16:14 发布 · 1.4k 阅读

1 ·

CC 4.0 BY-SA版权

dp-简单dp 同时被 2 个专栏收录

41 篇文章

订阅专栏

Hackerrank

8 篇文章

订阅专栏

本文介绍了一个经典的编程问题——最小糖果分配问题。问题中孩子们根据表现评分排队，表现更好的孩子需要获得比邻近孩子更多的糖果。文章提供了一种从两端向中间扫描的方法来确定每个孩子的糖果数量，以达到总糖果数最小的目标。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Candies

Problem Statement

Alice is a kindergarden teacher. She wants to give some candies to the children in her class. All the children sit in a line ( their positions are fixed), and each of them has a rating score according to his or her performance in the class. Alice wants to give at least 1 candy to each child. If two children sit next to each other, then the one with the higher rating must get more candies. Alice wants to save money, so she needs to minimize the total number of candies given to the children.

Input Format

The first line of the input is an integer N, the number of children in Alice's class. Each of the following N lines contains an integer that indicates the rating of each child.

1 <= N <= 105
1 <= ratingi <= 105

Output Format

Output a single line containing the minimum number of candies Alice must buy.

Sample Input

Sample Output

Explanation

Here 1, 2, 2 is the rating. Note that when two children have equal rating, they are allowed to have different number of candies. Hence optimal distribution will be 1, 2, 1.

题意：一堆小孩站成一排，每个小孩有一个分数，现在给各个小孩发糖果，每个小孩最少是一个糖果，如果这个小孩的分数比相邻小孩分数高，那么他得到的糖果也要比相邻小孩糖果多。问一共最少需要多少糖果。

从左到右扫一遍，从右到左扫一遍，求出每一个位置要求的最大值，相加得到结果。

代码：

#pragma warning(disable:4996)
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long ll;

#define INF 0x7fffffff
const int maxn = 100005;

int n, m;
int rating[maxn];
int dp1[maxn];
int dp2[maxn];

void input()
{
	int i;
	scanf("%d", &n);
	for (i = 1; i <= n; i++)
	{
		scanf("%d", &rating[i]);
		dp1[i] = 1;
		dp2[i] = 1;
	}
}

void solve()
{
	int i;
	for (i = 2; i <= n; i++)
	{
		if (rating[i] > rating[i - 1])
		{
			dp1[i] = dp1[i - 1] + 1;
		}
	}
	for (i = n-1; i >=1 ; i--)
	{
		if (rating[i] > rating[i + 1])
		{
			dp2[i] = dp2[i + 1] + 1;
		}
	}
	ll res = 0;
	for (i = 1; i <= n; i++)
	{
		res += max(dp1[i], dp2[i]);
	}
	cout << res;
}

int main()
{
	//freopen("i.txt","r",stdin);
	//freopen("o.txt","w",stdout);

	input();
	solve();

	//system("pause");
	return 0;
}