Problem G. Travelling Salesman Problem Input file: standard input Output file: standard output Time limit: 1 second Memory limit: 64 mebibytes 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 First line of the input contains one integer T (1 ≤ T ≤ 130) — number of test cases. For each test case, the first line contains two numbers n and 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). Examples standard input standard output 1 3 3 2 3 3 3 3 3 3 3 2 25 RRDLLDRR
题目分析：首先n,m有一个是奇数的时候，是可以遍历掉每一个格子的，全是偶数的之后 期望最好的情况是只留下一个不走。
小范围暴力找可以不走哪些格子，发现画出4*4的只能不走奇偶不同的。
如果全为偶数，考虑奇偶性，找最小的奇偶不同的点，不走。
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int MAXN = 100+10; 5 int a[MAXN][MAXN]; 6 7 void firstRow(int M) 8 { 9 for(int i=0;i<M-1;i++) printf("R"); 10 printf("D"); 11 for(int i=0;i<M-1;i++) printf("L"); 12 printf("D"); 13 } 14 15 void lastRow(int M) 16 { 17 printf("D"); 18 for(int i=0;i<M-1;i++) printf("L"); 19 printf("D"); 20 for(int i=0;i<M-1;i++) printf("R"); 21 } 22 23 void firstCol() 24 { 25 printf("D"); 26 printf("R"); 27 printf("U"); 28 printf("R"); 29 } 30 31 void lastCol() 32 { 33 printf("R"); 34 printf("U"); 35 printf("R"); 36 printf("D"); 37 } 38 39 void Printf(int M,int N,int m,int n) 40 { 41 int i=0,j=0; 42 while(i<(m/2)*2) firstRow(N),i+=2; 43 while(j<(n/2)*2) firstCol(),j+=2; 44 if(m%2==0) printf("DR"); 45 else printf("RD"); 46 j+=2; 47 while(j<N) lastCol(),j+=2; 48 i+=2; 49 while(i<M) lastRow(N),i+=2; 50 } 51 52 void Run() 53 { 54 int M,N; 55 int sum=0; 56 scanf("%d%d",&M,&N); 57 for(int i=0;i<M;i++) 58 for(int j=0;j<N;j++) 59 { 60 scanf("%d",&a[i][j]); 61 sum+=a[i][j]; 62 } 63 if(M%2==1) 64 { 65 printf("%d ",sum); 66 for(int i=0;i<M;i++) 67 { 68 for(int j=0;j<N-1;j++) 69 { 70 if(i%2==0) printf("R"); 71 else printf("L"); 72 } 73 if(i!=M-1) printf("D"); 74 } 75 return; 76 } 77 if(N%2==1) 78 { 79 printf("%d ",sum); 80 for(int j=0;j<N;j++) 81 { 82 for(int i=0;i<M-1;i++) 83 { 84 if(j%2==0) printf("D"); 85 else printf("U"); 86 } 87 if(j!=N-1) printf("R"); 88 } 89 return; 90 } 91 92 int Min=10000+1; 93 int rei,rej; 94 for(int i=0;i<M;i++) 95 { 96 for(int j=0;j<N;j++) 97 { 98 if((i%2+j%2)==1) 99 { 100 if(Min>a[i][j]) 101 { 102 Min=a[i][j];rei=i;rej=j; 103 } 104 } 105 } 106 } 107 printf("%d ",sum-Min); 108 Printf(M,N,rei,rej); 109 110 111 } 112 113 int main() 114 { 115 int T; 116 scanf("%d",&T); 117 while(T--) 118 { 119 Run(); 120 printf(" "); 121 } 122 return 0; 123 }