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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"
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)
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
題意：
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
題意：
有n戶人，打算把他們的房子圖上顏色，有red、green、blue三種顏色，每家人塗不同的顏色要花不同的費用。
並且相鄰兩戶人家之間的顏色要不同。求最小的總花費費用。
思路：
這個題與劉汝佳的算法競賽與入門經典中的數字三角形有點類似，能夠參照其方法。利用動態規劃的思想。找出
狀態轉移方程，dp[i][j%3+1] = a[i][j%3+1] + min(dp[i-1][(j-1)%3+1], dp[i-1][(j+1)%3+1]，這樣答案就出來了。
代碼：
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int dp[23][4];
int a[23][4];
int main()
{
int n, t;
scanf("%d",&t);
for(int k = 1; k <= t; k++)
{
scanf("%d",&n);
memset(a, 0, sizeof(a));
memset(dp, 0, sizeof(dp));
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= 3; j++)
scanf("%d",&a[i][j]);
}
for (int i = 1; i <= n; i++)
{
for (int j = 3; j < 6; j++)
{
dp[i][j%3+1] = a[i][j%3+1] + min(dp[i-1][(j-1)%3+1], dp[i-1][(j+1)%3+1]);
}
}
printf("Case %d: %d\n", k, min(dp[n][1], min(dp[n][2], dp[n][3])));
}
return 0;
}
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