lightoj-1047-Neighbor House【矩阵DP】

最新推荐文章于 2019-02-11 14:50:55 发布

Somethingwll

最新推荐文章于 2019-02-11 14:50:55 发布

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分类专栏： DP

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解决Mohammadpur镇居民给房子涂上红、绿、蓝三色的问题，确保相邻房子颜色不同，并求得最低总成本。使用动态规划算法，记录每种颜色的最小花费。

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题目链接：点击打开链接

1047 - Neighbor House

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Time Limit: 0.5 second(s)

Memory Limit: 32 MB

The people of Mohammadpur have decided to paint each of their houses red, green, or blue. They've also decided that no two neighboring houses will be painted the same color. The neighbors of house i are houses i-1 and i+1. The first and last houses are not neighbors.

You will be given the information of houses. Each house will contain three integers "R G B" (quotes for clarity only), where R, G and B are the costs of painting the corresponding house red, green, and blue, respectively. Return the minimal total cost required to perform the work.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case begins with a blank line and an integer n (1 ≤ n ≤ 20) denoting the number of houses. Each of the next n lines will contain 3 integers "R G B". These integers will lie in the range [1, 1000].

Output

For each case of input you have to print the case number and the minimal cost.

Sample Input

Output for Sample Input

13 23 12

77 36 64

44 89 76

31 78 45

26 40 83

49 60 57

13 89 99

Case 1: 137

Case 2: 96

题解：想那么长时间啊ZZ，就是一道简单的矩阵 DP 刚开始这都没看出来，好长时间没做过DP了，dp [ i ] [ j ] = k 表示第 i 行涂 j 这种颜色的最小花费为 k

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
int val[50][3],dp[50][3];
int main()
{
	int t,text=0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
			scanf("%d %d %d",&val[i][0],&val[i][1],&val[i][2]);
		memset(dp,0,sizeof(dp));
		for(int i=1;i<=n;i++)
		{
			dp[i][0]=val[i][0]+min(dp[i-1][1],dp[i-1][2]);
			dp[i][1]=val[i][1]+min(dp[i-1][0],dp[i-1][2]);
			dp[i][2]=val[i][2]+min(dp[i-1][0],dp[i-1][1]);
		}
		printf("Case %d: %d\n",++text,min(dp[n][0],min(dp[n][1],dp[n][2])));
	}
	return 0;
}