light oj 1047 - Neighbor House 动态规划

题目链接

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.………………

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，而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：

     动态规划，这个和刘汝佳算法竞赛入门经典P158的数字三角形有些相似，不过是求最小的值，而且有些限制，每次走到点和上次走的点不在同一列。

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]);
     
    
   

    

     

    
    

     
                //这里我用这种方法避免了分情况讨论，、减少了代码量
     
    
   

    

     

    
    

     

                  }
     
    
   

    

     

    
    

     

              }
     
    
   

    

     

    
    

     
        printf("Case %d: %d\n", k, min(dp[n][1], min(dp[n][2], dp[n][3])));
     
    
   

    

     

    
    

     

          }
     
    
   

    

     

    
    

     
    return 0;
     
    
   

    

     

    
    

     

      }
     
    
  
 

文章来源: xindoo.blog.csdn.net，作者：xindoo，版权归原作者所有，如需转载，请联系作者。

原文链接：xindoo.blog.csdn.net/article/details/8832898