hdu 5402 Travelling Salesman Problem（大模拟）
Problem Description
Teacher Mai is in a maze with n rows and m columns. There is a non-negative number in each cell. Teacher Mai wants to walk from the top left corner (1,1) to the bottom right corner (n,m). He can choose one direction and walk to this adjacent cell. However, he can't go out of the maze, and he can't visit a cell more than once. Teacher Mai wants to maximize the sum of numbers in his path. And you need to print this path.
Input
There are multiple test cases. For each test case, the first line contains two numbers n,m(1≤n,m≤100,n∗m≥2). In following n lines, each line contains m numbers. The j-th number in the i-th line means the number in the cell (i,j). Every number in the cell is not more than 104.
Output
For each test case, in the first line, you should print the maximum sum. In the next line you should print a string consisting of "L","R","U" and "D", which represents the path you find. If you are in the cell (x,y), "L" means you walk to cell (x,y−1), "R" means you walk to cell (x,y+1), "U" means you walk to cell (x−1,y), "D" means you walk to cell (x+1,y).
Sample Input
3 3
2 3 3
3 3 3
3 3 2
Sample Output
25
RRDLLDRR
Author
xudyh
Source
真是个鸟题，一开始把情况都考虑了，交上去直接WA了，后来意识到是n、m同为偶数有错。 其实只要找找规律就可以得出找的最小值的横纵坐标的和要为奇数，不过这个的顺序也是很难写啊。。。。。。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 int n,m; 6 int mp[106][106]; 7 int x,y; 8 int minn; 9 void get() 10 { 11 x = 1; y = 2; 12 for (int i = 1; i <= n;i++) 13 for (int j = 1; j <= m; j++) 14 if (((i + j) & 1) && mp[x][y] > mp[i][j]) x = i, y = j; 15 } 16 int main() 17 { 18 while(scanf("%d%d",&n,&m)==2) 19 { 20 int sum=0; 21 for(int i=1;i<=n;i++) 22 { 23 for(int j=1;j<=m;j++) 24 { 25 scanf("%d",&mp[i][j]); 26 sum+=mp[i][j]; 27 } 28 } 29 if((n&1)) 30 { 31 32 printf("%d\n",sum); 33 if(n==1) 34 { 35 for(int i=1;i<m;i++) 36 printf("R"); 37 printf("\n"); 38 continue; 39 } 40 for(int i=1;i<m;i++) 41 printf("R"); 42 printf("D"); 43 for(int i=2;i<=n;i++) 44 { 45 if(i%2==0) 46 { 47 for(int j=1;j<m;j++) 48 printf("L"); 49 printf("D"); 50 } 51 else 52 { 53 if(i!=n) 54 { 55 for(int i=1;i<m;i++) 56 printf("R"); 57 printf("D"); 58 } 59 else 60 { 61 for(int i=1;i<m;i++) 62 printf("R"); 63 } 64 } 65 } 66 printf("\n"); 67 68 } 69 else if((m%2)) 70 { 71 printf("%d\n",sum); 72 if(m==1) 73 { 74 for(int i=1;i<n;i++) 75 printf("D"); 76 printf("\n"); 77 continue; 78 } 79 for(int i=1;i<n;i++) 80 printf("D"); 81 printf("R"); 82 for(int i=2;i<=m;i++) 83 { 84 if(i%2==0) 85 { 86 for(int j=1;j<n;j++) 87 printf("U"); 88 printf("R"); 89 } 90 else 91 { 92 if(i!=m) 93 { 94 for(int j=1;j<n;j++) 95 printf("D"); 96 printf("R"); 97 } 98 else 99 { 100 for(int j=1;j<n;j++) 101 printf("D"); 102 } 103 } 104 } 105 printf("\n"); 106 } 107 else if((m%2)==0 && (n%2)==0) 108 { 109 get(); 110 printf("%d\n", sum - mp[x][y]); 111 for (int i = 1; i <= n; i += 2) 112 { 113 if (x == i || x == i + 1) 114 { 115 for (int j = 1; j < y; j++) 116 { 117 if (j & 1) printf("D"); else printf("U"); 118 printf("R"); 119 } 120 if (y < m) printf("R"); 121 for (int j = y + 1; j <= m; j++) 122 { 123 if (j & 1) printf("U"); else printf("D"); 124 if (j < m) printf("R"); 125 } 126 if (i < n - 1) printf("D"); 127 } 128 else if (i < x) 129 { 130 for (int j = 1; j < m; j++) printf("R"); 131 printf("D"); 132 for (int j = 1; j < m; j++) printf("L"); 133 printf("D"); 134 } 135 else 136 { 137 for (int j = 1; j < m; j++) printf("L"); 138 printf("D"); 139 for (int j = 1; j < m; j++) printf("R"); 140 if (i < n - 1) printf("D"); 141 } 142 } 143 printf("\n"); 144 } 145 146 147 } 148 return 0; 149 }
