LightOJ 1047 Neighbor House （DP 数字三角形变形）

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本文探讨了在Mohammadpur地区进行房屋着色的问题，要求相邻房屋不能使用相同的颜色。通过动态规划方法解决该问题，给出求解最小总成本的方法，并提供了一个样例输入输出以验证算法的有效性。

1047 - Neighbor House

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Time Limit: 0.5 second(s)

Memory Limit: 32 MB

The people of Mohammadpur have decided to paint each oftheir houses red, green, or blue. They've also decided that no two neighboringhouses will be painted the same color. The neighbors of housei arehouses i-1 and i+1. The first and last houses are not neighbors.

You will be given the information of houses. Each house willcontain three integers"R G B" (quotes for clarity only),where R, G and B are the costs of painting the correspondinghouse red, green, and blue, respectively. Return the minimal total costrequired 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 integern (1≤ n ≤ 20) denoting the number of houses. Each of the next nlines will contain 3 integers "R G B". These integers will liein the range[1, 1000].

Output

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

Sample Input

Output for Sample Input

13 23 12

77 36 64

44 89 76

31 78 45

26 40 83

49 60 57

13 89 99

Case 1: 137

Case 2: 96

题目链接：http://lightoj.com/volume_showproblem.php?problem=1047

题目大意：相邻行不能取同一列的数字，从上往下路径最大权值和问题。

解题思路：只有三列，每步有两次选择，dp[i][j]表示到第a[i][j]步的最小数字累加和，起点处分三条路，最后选出和最小的一条路。

代码如下：

#include <cstdio>
#include <cstring>
int a[25][4],dp[25][4];
int min2(int a,int b)
{
    return a>b?b:a;
}
int min3(int a,int b,int c)
{
    if(a<=b&&a<=c)
        return a;
    if(b<=a&&b<=c)
        return b;
    if(c<=b&&c<=a)
        return c;
}
int main()
{
    int t,n,i,j,cnt=0;
    scanf("%d",&t);
    while(t--)
    {
        int ans;
        memset(dp,0,sizeof(dp));
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        for(i=0;i<n;i++)
            for(j=0;j<3;j++)
                scanf("%d",&a[i][j]);
        dp[0][0]=a[0][0];
        dp[0][1]=a[0][1];
        dp[0][2]=a[0][2];
        for(i=1;i<n;i++)
        {
            dp[i][0]=a[i][0]+min2(dp[i-1][2],dp[i-1][1]);
            dp[i][1]=a[i][1]+min2(dp[i-1][2],dp[i-1][0]);
            dp[i][2]=a[i][2]+min2(dp[i-1][0],dp[i-1][1]);
        }
        ans=min3(dp[n-1][0],dp[n-1][1],dp[n-1][2]);
        printf("Case %d: %d\n",++cnt,ans);
    }
    return 0;
}