HDU 4283 You are the one（区间DP）

最新推荐文章于 2021-10-20 16:48:09 发布

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最新推荐文章于 2021-10-20 16:48:09 发布

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分类专栏：动态规划文章标签： ACM dp

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本文介绍了一种利用区间动态规划（DP）解决排位问题的方法，具体场景为在小型演播室中，导演需要调整男生们的出场顺序，以最小化因等待导致的不满情绪。通过详细解析算法步骤和代码实现，展示了如何通过动态规划优化问题解决方案。

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题目大意：

　　The TV shows such as You Are the One has been very popular. In order to meet the need of boys who are still single, TJUT hold the show itself. The show is hold in the Small hall, so it attract a lot of boys and girls. Now there are n boys enrolling in. At the beginning, the n boys stand in a row and go to the stage one by one. However, the director suddenly knows that very boy has a value of diaosi D, if the boy is k-th one go to the stage, the unhappiness of him will be (k-1)*D, because he has to wait for (k-1) people. Luckily, there is a dark room in the Small hall, so the director can put the boy into the dark room temporarily and let the boys behind his go to stage before him. For the dark room is very narrow, the boy who first get into dark room has to leave last. The director wants to change the order of boys by the dark room, so the summary of unhappiness will be least. Can you help him?

解题思路：

区间DP，dp[i][j]表示从i到j的沮丧值，枚举第i个人的出场顺序。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define LL long long 
using namespace std;
const int MAXN = 100 + 10;
const int inf = 0x3f3f3f3f;
int dp[MAXN][MAXN];
int a[MAXN], sum[MAXN];
int N;
int main()
{
	int T, kcase = 1;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d", &N);
		memset(sum, 0, sizeof(sum));
		memset(dp, 0, sizeof(dp));
		for(int i=1;i<=N;i++)
		{
			scanf("%d", &a[i]);
			sum[i] = sum[i-1] + a[i];
		}
		memset(dp, 0, sizeof(dp));
		for(int i=1;i<=N;i++)
		{
			for(int j=i+1;j<=N;j++)
				dp[i][j] = inf;
		}
		for(int len=1;len<N;len++)
		{
			for(int i=1;i+len<=N;i++)
			{
				int j = i + len;
				for(int k=1;k<=j-i+1;k++)//第i个人第K个上场
				{
					dp[i][j] = min(dp[i][j], dp[i+1][i+k-1] + a[i] * (k-1) + dp[i+k][j] + k * (sum[j] - sum[i+k-1]));
					/*dp[i+1][i+k-1]表示前k-1个人的沮丧值，a[i] * (k-1）表示第i个人的沮丧值，而i+k到j的这些人由于出场位置都增加了K，所以总的沮丧值增加了k * (sum[j] - sum[i+k-1]）；*/
				}
			}
		}
		printf("Case #%d: %d\n", kcase++, dp[1][N]);
	}
	return 0;
}