1220.Look for homework

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Description

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 disturded.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(2<=n,m<=10),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 30011001103 3000000000

Sample Output

4RDRD4DDRR
就是一个迷宫问题，然后找最短的路线，然后如果有多个最短路线就按字典序输出最靠前的。。当时做一脸**，后来想了想，让他一直按字典序查找直到出口的最短路线就是答案了。。。。就直接bfs（因为一些语法错误一直WA。。。细心，细心！）
#include<cstdio>#include<cstring>using namespace std;long long a,b,n,m;struct bbq{    int x,y,pre;    char ch;};bbq p[105];int vis[13][13];int dir[4][2]= {{1,0},{0,-1},{0,1},{-1,0}};char s[4]= {'D','L','R','U'};int cmp(int x,int y){    if(x<0||x>n-1||y<0||y>m-1||vis[x][y])        return 0;    return 1;}int serch(int x,int y){    int xx,yy;    a=0;    b=1;    p[a].x=x;    p[a].y=y;    p[a].pre=-1;    vis[x][y]=1;    while(a<b)    {        for(int i=0; i<=3; i++)        {            //printf("1");            xx=p[a].x+dir[i][0];            yy=p[a].y+dir[i][1];            /*printf("x=%d y=%d ",p[a].x,p[a].y,a,b);            printf("a=%d ",a);            printf("b=%d\n",b);*/            if(cmp(xx,yy))            {                vis[xx][yy]=1;                p[b].x=xx;                p[b].y=yy;                p[b].pre=a;                p[b].ch=s[i];                b++;            }            if(xx==n-1&&yy==m-1)            {                return b-1;            }        }        a++;    }}char ans[100];int main(){    while(scanf("%d %d",&n,&m)!=EOF)    {        memset(vis,0,sizeof(vis));        int t,b,v=0;        char ss[12][12];        for(int i=0; i<n; i++)        {            getchar();            scanf("%s",ss[i]);            for(int j=0; j<m; j++)            {                if(ss[i][j]=='1')                    vis[i][j]=1;                else vis[i][j]=0;            }        }        b=serch(0,0);        while(p[b].pre!=-1)        {            ans[v]=p[b].ch;            v++;            b=p[b].pre;        }        printf("%d\n",v);        for(int i=v-1;i>=0;i--)        printf("%c",ans[i]);        printf("\n");    }}

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