light oj1074

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 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].

For each case of input you have to print the case number and the minimal cost.

2

4

13 23 12

77 36 64

44 89 76

31 78 45

3

26 40 83

49 60 57

13 89 99

Case 1: 137

Case 2: 96

Use simple DP

题意：

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。

并且相邻两户人家之间的颜色要不同。求最小的总花费费用。

思路：

这个题与刘汝佳的算法竞赛与入门经典中的数字三角形有点类似，能够參照其方法。利用动态规划的思想。找出

状态转移方程，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]，这样答案就出来了。

代码：

#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;
}