蓝桥杯 算法训练 ALGO-36 传纸条

 1 #include <iostream>
 2 #include<stdio.h>
 3 #include<cmath>
 4 using namespace std;
 5 
 6 #define MAX_NUM 52
 7 
 8 int map[MAX_NUM][MAX_NUM];     //好心程度 | 权值
 9 int dp[MAX_NUM][MAX_NUM][MAX_NUM][MAX_NUM];
10 
11 int maxPath(int m, int n)
12 {
13     for (int x1 = 1; x1 <= m; x1++)
14     {
15         for (int y1 = 1; y1 <= n; y1++)
16         {
17             for (int x2 = 1; x2 <= m; x2++)
18             {
19                 for (int y2 = 1; y2 <= n; y2++)
20                 {
21                     /*
22                         如果第一个人没有走到最后一行或最后一列，并且两个人没有重复 
23                         因为走到最后一行或最后一列，容易造成第二个人无路可走的情况 
24                     */
25                     if ((x1 < m || y1 < n) && x1 == x2 && y1 == y2)     
26                     {
27                         continue;
28                     }
29                     dp[x1][y1][x2][y2] = max( max(dp[x1-1][y1][x2-1][y2], dp[x1-1][y1][x2][y2-1]), 
30                                               max(dp[x1][y1-1][x2-1][y2], dp[x1][y1-1][x2][y2-1]))
31                                              + map[x1][y1] + map[x2][y2];
32                 }
33             }
34         }
35     }
36     return dp[m][n][m][n]; 
37 }
38 
39 int main()
40 {
41     int m, n;
42     scanf("%d%d", &m, &n);
43     
44     for (int i = 1;i <= m; i++)
45         for (int j = 1;j <= n; j++)
46             scanf("%d", &map[i][j]);
47         
48     int ans = maxPath(m, n);
49     printf("%d\n", ans);
50         
51     return 0;
52 }

 1 #include <iostream>
 2 #include<cmath>
 3 using namespace std;
 4 
 5 #define MAX_NUM 52
 6 
 7 int map[MAX_NUM][MAX_NUM];     //好心程度 | 权值
 8 int dp[MAX_NUM+MAX_NUM][MAX_NUM][MAX_NUM];
 9 
10 int maxPath(int m, int n)
11 {
12     for (int k = 1;k <= m+n-3; k++)
13     {
14         for (int x1 = 0; x1 <= k; x1++)
15         {
16             for (int x2 = 0; x2 <= k; x2++)
17             {
18                 if (x1 == x2)    //x1 == x2 相当于(x1 == x2 && y1 = y2) 
19                 {
20                     continue;
21                 }
22                 dp[k][x1][x2] = max(max(dp[k-1][x1][x2], dp[k-1][x1-1][x2-1]),
23                                     max(dp[k-1][x1-1][x2], dp[k-1][x1][x2-1]))
24                                 + map[x1][k-x1] + map[x2][k-x2];
25             }
26         }
27     }
28     return dp[m+n-3][m-1][m-2];
29 }
30 
31 int main()
32 {
33     int m, n;
34     scanf("%d%d", &m, &n);
35     
36     for (int i = 0; i < m; i++)
37         for (int j = 0; j < n; j++)
38             scanf("%d", &map[i][j]);
39             
40     int ans = maxPath(m, n);
41     printf("%d\n", ans);
42         
43     return 0;
44 }

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath> 
 4 using namespace std;
 5  
 6 #define MAX_NUM 52
 7 
 8 int map[MAX_NUM][MAX_NUM];   //好心程度 | 权值 
 9 int dp[MAX_NUM][MAX_NUM];
10  
11 int maxPath(int m, int n)  
12 {  
13       
14     for (int k = 1; k <= m+n-3; k++)  
15     {  
16         for (int x1 = m-1; x1 >= 0; x1--)  
17         {  
18             for (int x2 = m-1; x2 > x1; x2--)  
19             {  
20                 if ( k >= x1 && k >= x2)    //x + y = k,当k >= x时，说明还在矩阵范围之内  
21                 {
22                     dp[x1][x2] = max(max(dp[x1][x2], dp[x1-1][x2-1]), 
23                                      max(dp[x1-1][x2], dp[x1][x2-1])) 
24                                  + map[x1][k-x1] + map[x2][k-x2];
25                 }
26             }  
27         }  
28     }  
29     return dp[m-2][m-1]; 
30 }
31   
32 int main()  
33 {
34     int m, n;
35     scanf("%d %d", &m, &n);
36       
37     for (int i = 0;i < m; i++)  
38         for (int j = 0; j < n; j++)  
39             scanf("%d", &map[i][j]);  
40     
41     int ans = maxPath(m, n);
42     printf("%d\n", ans); 
43     
44     return 0;  
45 }