LIGHT OJ 1047 - Neighbor House 【DP数塔】

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
2   4 13 23 12 77 36 64 44 89 76 31 78 45   3 26 40 83 49 60 57 13 89 99	Case 1: 137 Case 2: 96

题意：有一排房子，每个房子可以染成三种颜色，相邻的房子不能染成相同的颜色，给出每个房子染成每种颜色的价钱，求将所有房子染色的最小花费；

思路：n最大20，枚举的话是3*2^19时间受不了，就用DP来优化呗，造成时间复杂度高的原因就是每一个中间的状态都查询了多次，为了避免这种情况，我们应该将中间状态保存起来，使每个中间结果只·计算一次：dp[i][j]:表示涂到第i房子第j种颜色的最低花费，来个循环，最后选个最小值就行了；

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int a[1111][7],dp[1111][6];

int main()
{
	int T,N,i,j,t1,t2,Kase=0;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&N); memset(a,0,sizeof(a));
		for(i=1;i<=N;++i)
		{
			for(j=1;j<=3;++j)
			{
				scanf("%d",&a[i][j]);
			}
		}
		for(i=1;i<=N;++i)
		{
			for(j=1;j<=3;++j)
			{
				if(j==1) t1=2,t2=3;
				else if(j==2) t1=1,t2=3;
				else t1=1,t2=2;
				dp[i][j]=a[i][j]+min(dp[i-1][t1],dp[i-1][t2]);
			}
		}
		printf("Case %d: %d\n",++Kase,min(dp[N][1],min(dp[N][2],dp[N][3])));
	}
	return 0;
}