Description

(the picture has no relation with this problem…I just want to add a picture. emmm…)

Super scholar robs the all homework of others. The monitor decides to combat with the super scholar so as to help students to get back the homework. But super scholar lives in a castle because he doesn’t want to be disturbed. The outside of the castle is a maze with two dimension grids. Entering the castle must pass the maze. The monitor needs to save time because of accompanying his girlfriend, so he wants to make a student named Huachao Wei help him who hasn’t a girlfriend. Now he has the map of the maze and needs to calculate the shortest path.
Input
The input consists several cases.For each case starts with two integers n,m(​),symbolizing the length and width of the maze.The next n lines contain m numbers with no space and value for only 0 and 1.The essence is that 0 can pass ,1 can’t pass. Now you are in the place of (1,1) refering to the top left corner.the export is in the place of (n,m).Each step can only walk with up,down,left and right.
Output
For each case,the first line prints the least number of steps to arrive the castle.The second line prints k characters for U,D,L,R,representing up,down,left,right.if there are many paths with the same length,please print the path with Minimum dictionary.It is guarantees that there must have a path for arriving the castle.
Sample Input
3 3
001
100
110
3 3
000
000
000
Sample Output
4
RDRD
4
DDRR

#include <cstdio>
#include <iostream>
#include <queue>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

///保留路径不允许元素消失
///queue是删除性访问不符合条件
///所以用数组代替（省内存的事不考虑了）
struct node
{
    int x, y, pre;
    char s;
} a[105];

///图的界限
int n, m;
///移动方向（坐标系逆转90°）
int dx[] = { 1, 0, 0, -1};
int dy[] = { 0, -1, 1, 0};
char dir[] = {'D', 'L', 'R', 'U' };
///图的存储
string mp[11];
///访问标记
bool vis[11][11];
///路径存储
char c[105];
///搜索引导
int fro, bac;

void print(int i, int j);

void bfs(int x, int y)
{
    fro = 0;
    bac = 1;
    a[fro].x = x;
    a[fro].y = y;
    a[fro].pre = -1;
    vis[x][y] = 1;
    while(fro < bac)///小小的设计体现了队列的先进先出、穷竭搜索！
    {
        for(int i = 0; i < 4; ++i)
        {
            int temx = a[fro].x + dx[i];
            int temy = a[fro].y + dy[i];
            if(temx < 0 || temy < 0 || temx >= n || temy >= m || vis[temx][temy])
                continue;
            vis[temx][temy] = 1;
            a[bac].x = temx;
            a[bac].y = temy;
            a[bac].pre = fro;
            a[bac].s = dir[i];
            bac++;

            if(temx == n - 1 && temy == m - 1)
            {
                print(fro, bac - 1);
                return ;
            }
        }
        fro++;
    }
    return ;
}

void print(int i, int j)
{
    int tem = i;
    int len = 0;
    c[len++] = a[j].s;
    while(a[tem].pre != -1)
    {
        c[len++] = a[tem].s;
        tem = a[tem].pre;
    }
    cout << len << '\n';
    ///存了倒序的路径，自然要倒序输出
    for(int k = len - 1; k >= 0; --k)
    {
        cout << c[k];
    }
    cout << '\n';
}

int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        for(int i = 0; i < n; ++i)
        {
            cin >> mp[i];
            for(int j = 0; j < m; ++j)
            {
                if(mp[i][j] == '1' )
                    vis[i][j] = 1;
                ///不可走的地方直接标记
            }
        }
        bfs(0, 0);
        memset(vis, 0, sizeof(vis));///满足多组输入
    }
    return 0;
}