light oj 1047-neighbor house

ime Limit:500MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Description

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

Hint

Use simple DP

程序分析：此题的题意为有n户人，打算把他们的房子图上颜色，有red、green、blue三种颜色，每家人涂不同的颜色要花不同的费用，而且相邻两户人家之间的颜色要不同，求最小的总花费费用。我们所要使用的方法就是DP，其中有一个小问题值得注意就是开始我用C提交的时候就会提示CE，之后我们C++交的时候就过了，所以有时候选择语言也是很重要的。

程序代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
 
using namespace std;
 
int dp[][];
int a[][];
int min(int a,int b)
{
    if(a>b)
        return b;
    else return a;
}
int main()
{
    int n, t;
    scanf("%d",&t);
    for(int k = ; k <= t; k++)
    {
        scanf("%d",&n);
        memset(a, , sizeof(a));
        memset(dp, , sizeof(dp));
        for (int l= ; l <= n; l++)
        {
            for (int j = ; j <= ; j++)
                scanf("%d",&a[l][j]);
        }
        for (int i = ; i <= n; i++)
        {
            for (int j = ; j < ; j++)
            {
                dp[i][j%+] = a[i][j%+] + min(dp[i-][(j-)%+], dp[i-][(j+)%+]);
              //这里我用这种方法避免了分情况讨论，、减少了代码量
 
            }
        }
        printf("Case %d: %d\n", k, min(dp[n][], min(dp[n][], dp[n][])));
    }
    return ;
}