light oj 1047 - Neighbor House(贪心)

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

2

4

13 23 12

77 36 64

44 89 76

31 78 45

3

26 40 83

49 60 57

13 89 99

Sample Output

Case 1: 137

Case 2: 96

题解：这个题我一开始想的是每次都找最小的的那一个（除了第一个），并且颜色不相同，这时候开个标记数组，把每次可以涂得选一个较小的，这种解法有很大的问题，既然你考虑了第一个可能不是最小的那后边也可能不是最小的相加就是和最小的，因为你所致的最小是去除不能标记点的最小。而不是单纯每次求最小，和就是最小，是受前边影响的，那我们要求的是和最小，我们就可以每次找和最小的，这样最后找出来肯定是最小的。

附上两次的代码：

错误代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
	int n;
	int a[1005][3];
    int mark[1005][3];
	scanf("%d",&n);
	for(int t=0;t<n;t++)
	{
		memset(mark,0,sizeof(mark));
		int m;
		scanf("%d",&m);
		for(int j=0;j<m;j++)
		{
			for(int k=0;k<3;k++)
			{
				scanf("%d",&a[j][k]);
			}
		}
		long long sum1=a[0][0];
		memset(mark,0,sizeof(mark));
		mark[0][0]=1;
		int temp1;

	    for(int j=1;j<m;j++)
	    {
	    	int minn=9999;
	    	for(int k=0;k<3;k++)
	    	{
	    		if(!mark[j-1][k])
	    		{
	    			if(minn>a[j][k])
	    			{
	    				minn=a[j][k];
	    				temp1=k;
					}
				}

			}
			sum1+=minn;
			mark[j][temp1]=1;

		}
		long long sum2=a[0][1];
		memset(mark,0,sizeof(mark));
		mark[0][1]=1;
		int temp2;

	    for(int j=1;j<m;j++)
	    {
	    	int minn=9999;
	    	for(int k=0;k<3;k++)
	    	{
	    		if(!mark[j-1][k])
	    		{
	    			if(minn>a[j][k])
	    			{
	    				minn=a[j][k];
	    				temp2=k;
					}
				}

			}
			sum2+=minn;
			mark[j][temp2]=1;

		}
		long long sum3=a[0][2];
		memset(mark,0,sizeof(mark));
		mark[0][2]=1;
		int temp3;

	    for(int j=1;j<m;j++)
	    {
	    	int minn=9999;
	    	for(int k=0;k<3;k++)
	    	{
	    		if(!mark[j-1][k])
	    		{
	    			if(minn>a[j][k])
	    			{
	    				minn=a[j][k];
	    				temp3=k;
					}
				}

			}
			sum3+=minn;
			mark[j][temp3]=1;

		}
		if(min(sum1,sum2)>sum3)
		{
			printf("Case %d: %lld\n",t+1,sum3);
		}
		else
		{
			printf("Case %d: %lld\n",t+1,min(sum1,sum2));
		}
	}

	return 0;
} 

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

int main()
{
	int a[1005][3];
	int sum[1005][3];
	int n;
	scanf("%d",&n);
	for(int t=1;t<=n;t++)
	{
	   int m;
	   scanf("%d",&m);
	   for(int j=0;j<m;j++)
	   {
	   	for(int k=0;k<3;k++)
	   	{
	   	  scanf("%d",&a[j][k]);
		}
	   }
	   memset(sum,0,sizeof(sum));
	   sum[0][0]=a[0][0];
	   sum[0][1]=a[0][1];
	   sum[0][2]=a[0][2];
	   for(int  j=1;j<m;j++)
	   {
	   	sum[j][0]=min(sum[j-1][1],sum[j-1][2])+a[j][0];
	   	sum[j][1]=min(sum[j-1][0],sum[j-1][2])+a[j][1];
	   	sum[j][2]=min(sum[j-1][0],sum[j-1][1])+a[j][2];
	   }
	   printf("Case %d: %d\n",t,(min(sum[m-1][0],sum[m-1][1])<sum[m-1][2])?min(sum[m-1][0],sum[m-1][1]):sum[m-1][2]);
	}

	return 0;
}