HDU 5402 Travelling Salesman Problem

Problem Description

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

如果n和m里面有一个是奇数那么全部走遍就好了。

否则要找一个最小的点不要，这个点的坐标要满足x+y是奇数

如果不是的话，舍弃该点一定会导致另外一个点也走不到。

然后找到这个点，暴力的一行一行的跑就好了。

#pragma comment(linker, "/STACK:1024000000,1024000000") 
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const LL base = 1e9 + 7;
const int maxn = 105;
LL T, n, m, a[maxn][maxn], sum, x, y;

inline void read(int &x)
{
    char ch;
    while ((ch = getchar())<'0' || ch>'9');
    x = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9') x = x * 10 + ch - '0';
}

void get()
{
    x = 1;    y = 2;
    for (int i = 1; i <= n;i++)
        for (int j = 1; j <= m; j++) 
            if (((i + j) & 1) && a[x][y] > a[i][j]) x = i, y = j;
}

int main()
{
    while (scanf("%lld%lld", &n, &m) !=EOF)
    {
        sum = 0;
        for (int i = 1; i <= n;i++)
            for (int j = 1; j <= m; j++)
            {
                scanf("%lld", &a[i][j]);
                sum += a[i][j];
            }
        if (n & 1 || m & 1)
        {
            printf("%lld\n", sum);
            if (n & 1)
            {
                for (int i = 1; i <= n; i++)
                {
                    for (int j = 1; j < m; j++) 
                        if (i & 1) printf("R"); else printf("L");
                    if (i < n) printf("D"); else printf("\n");
                }
            }
            else
            {
                for (int i = 1; i <= m; i++)
                {
                    for (int j = 1; j < n; j++)
                        if (i & 1) printf("D"); else printf("U");
                    if (i < m) printf("R"); else printf("\n");
                }
            }
        }
        else
        {
            get();
            printf("%lld\n", sum - a[x][y]);
            for (int i = 1; i <= n; i += 2)
            {
                if (x == i || x == i + 1)
                {
                    for (int j = 1; j < y; j++)
                    {
                        if (j & 1) printf("D"); else printf("U");
                        printf("R");
                    }
                    if (y < m) printf("R");
                    for (int j = y + 1; j <= m; j++)
                    {
                        if (j & 1) printf("U"); else printf("D");
                        if (j < m) printf("R");
                    }
                    if (i < n - 1) printf("D");
                }
                else if (i < x)
                {
                    for (int j = 1; j < m; j++) printf("R");
                    printf("D");
                    for (int j = 1; j < m; j++) printf("L");
                    printf("D");
                }
                else
                {
                    for (int j = 1; j < m; j++) printf("L");
                    printf("D");
                    for (int j = 1; j < m; j++) printf("R");
                    if (i < n - 1) printf("D");
                }
            }
            printf("\n");
        }
    }
    return 0;
}