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 Bare 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
题目大意：给n座房子刷漆，每座房子可以刷红，绿，蓝色中的任何一种，粉刷每一种颜色都要花费一定的money，但是相邻两座房子的颜色不能相同，
要求输出最少的花费。用i表示1~n座房子，用j表示三种颜色的花费。
分析：先输入1~n的房子中每种颜色所需的花费p[i][j]，接着从1~n先比较第一座房子中所需花费最少的并标记其颜色，这其中有三种情况，j=1,2,3，
分别表示第一，二，三种颜色，用一个二维数组dp[i][j]=min(dp[i][j-2],dp[i][j-3])+p[i][j](当j=1时)，然后比较第二座房子所需的花费最少，判断其颜色是否与第一座相
同，不同则直接累加，否则比较出第二少的，再累加。
代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n;
int dp[25][5],p[25][5];
int main()
{
int t,i,j,kase=0;
scanf("%d",&t);
while(t--)
{
memset(dp,0,sizeof(dp));
scanf("%d",&n);
for(i=1;i<=n;i++)
for(j=1;j<=3;j++)
scanf("%d",&p[i][j]);
for(i=1;i<=n;i++)
for(j=1;j<=3;j++)
{
if(j==1)
dp[i][j]=min(dp[i-1][2],dp[i-1][3])+p[i][j];
if(j==2)
dp[i][j]=min(dp[i-1][1],dp[i-1][3])+p[i][j];
if(j==3)
dp[i][j]=min(dp[i-1][1],dp[i-1][2])+p[i][j];
}
printf("Case %d: %d
",++kase,min(dp[n][1],min(dp[n][2],dp[n][3])));
}
return 0;
}