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
Hint
Use simple DP
題解： 有n家人，要把他們的房子塗上顏色，有紅、綠、藍三種顏色，每家塗不同的顏色要花不同的費用，而且相鄰兩戶人家之間的顏色要不同，求最小的總花費費用。
看案例可以從第一行往下模擬，然後就知道了怎麼去實現，還是動態規劃，求最小的值，有限制條件，每次走到點和上次走的點不在同一列。
代碼:
#include<iostream>
#include<cstdio>
using namespace std;
int dp[1005][1005];
int a[1005][1005];
int main()
{
int T,n,k=1;
cin>>T;
while(T--)
{
cin>>n;
for(int i=1; i<=n; i++)
for(int j=1; j<=3; j++)
cin>>a[i][j];
for(int i=1; i<=n; i++)
for(int j=1; j<=3; j++)
{
dp[i][j%3+1]=min(dp[i-1][(j-1)%3+1],dp[i-1][(j+1)%3+1])+a[i][j%3+1]; // 相當於滾筒數組，優化
}
int total=min(dp[n][1],min(dp[n][2],dp[n][3]));
cout<<"Case "<<k++<<": "<<total<<endl;
}
}
