Travelling Salesman Problem
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 829 Accepted Submission(s): 182
Special Judge
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
1 #include <stdio.h>
2 #include <string.h>
3 int main()
4 {
5 int n,m;
6 int a[106][106],s,x,y;
7 int i,j,k;
8 while(scanf("%d %d",&n,&m)!=EOF)
9 {
10 s=0;
11 int mi;
12 for(i=1;i<=n;i++)
13 {
14 for(j=1;j<=m;j++)
15 {
16 scanf("%d",&a[i][j]);
17 s=s+a[i][j];
18 }
19 }
20 if(n%2==1)
21 {
22 printf("%d\n",s);
23 for(i=1;i<=n;i++)
24 {
25 if(i%2==1)
26 {
27 for(j=1;j<m;j++)
28 printf("R");
29 }
30 else
31 {
32 for(j=1;j<m;j++)
33 printf("L");
34 }
35 if(i!=n)
36 printf("D");
37 }
38 printf("\n");
39 }
40 else if(m%2==1)
41 {
42 printf("%d\n",s);
43 for(j=1;j<=m;j++)
44 {
45 if(j%2==1)
46 {
47 for(i=1;i<n;i++)
48 printf("D");
49 }
50 else
51 {
52 for(i=1;i<n;i++)
53 printf("U");
54 }
55 if(j!=m)
56 printf("R");
57 }
58 printf("\n");
59 }
60 else
61 {
62 mi=a[1][2];
63 x=1,y=2;
64 //printf("ok\n");
65 for(i=1;i<=n;i++)
66 {
67 for(j=1;j<=m;j++)
68 {
69 if(i%2==1 && j%2==0 && a[i][j]<mi)
70 {
71 mi=a[i][j];
72 x=i,y=j;
73 }
74 if(i%2==0 && j%2==1 && a[i][j]<mi)
75 {
76 mi=a[i][j];
77 x=i,y=j;
78 }
79 }
80 }
81 s=s-mi;
82 printf("%d\n",s);
83 for(i=1;2*i<x;i++)
84 {
85 for(j=1;j<m;j++)
86 {
87 printf("R");
88 }
89 printf("D");
90 for(j=1;j<m;j++)
91 {
92 printf("L");
93 }
94 printf("D");
95 }
96 //printf("ok\n");
97 int flg=1;
98 for(i=1;i<=m;i++)
99 {
100 if(y==i)
101 {
102 if(i==m && x!=n-1)
103 {
104 printf("D");
105 continue;
106 }
107 if(i!=m)
108 printf("R");
109 }
110 else
111 {
112 if(i==m)
113 {
114 if(x==n-1 || x==n)
115 printf("D");
116 else
117 printf("DD");
118 }
119 else if(flg==1)
120 {
121 printf("DR");flg=2;
122 }
123 else if(flg==2)
124 {
125 printf("UR");flg=1;
126 }
127 }
128 }
129 //printf("ok\n");
130 if(x%2==1)
131 x++;
132 for(i=x+1;i<=n;i++)
133 {
134 if(i%2==1)
135 {
136 for(j=1;j<m;j++)
137 printf("L");
138 }
139 else
140 {
141 for(j=1;j<m;j++)
142 printf("R");
143 }
144 if(i!=n)
145 printf("D");
146 }
147 printf("\n");
148 }
149 }
150 return 0;
151 }