The following figure shows a plan view of a maze, is a disorder in which the mark 1, the mark to be 0 in
place at the prevailing.
010000
000100
001001
110000
entrance to the upper left corner of the maze, export to the lower right corner, in the maze, which can only come from a position of
one of its upper and lower, one left, and right directions.
For the above maze, starting from the inlet, it can be by sequentially DRRURRDDDR maze,
a total of 10 steps. Wherein D, U, L, R respectively down, up, left, right away.
For more complex maze following (line 30 50), please find a way through the maze,
the number of steps which uses the minimum, the minimum number of steps in the premise, find a lexicographically smallest as the answer.
Note that D <L <R <U in the lexicographic order.
Input:
01010101001011001001010110010110100100001000101010 00001000100000101010010000100000001001100110100101 01111011010010001000001101001011100011000000010000 01000000001010100011010000101000001010101011001011 00011111000000101000010010100010100000101100000000 11001000110101000010101100011010011010101011110111 00011011010101001001001010000001000101001110000000 10100000101000100110101010111110011000010000111010 00111000001010100001100010000001000101001100001001 11000110100001110010001001010101010101010001101000 00010000100100000101001010101110100010101010000101 11100100101001001000010000010101010100100100010100 00000010000000101011001111010001100000101010100011 10101010011100001000011000010110011110110100001000 10101010100001101010100101000010100000111011101001 10000000101100010000101100101101001011100000000100 10101001000000010100100001000100000100011110101001 00101001010101101001010100011010101101110000110101 11001010000100001100000010100101000001000111000010 00001000110000110101101000000100101001001000011101 10100101000101000000001110110010110101101010100001 00101000010000110101010000100010001001000100010101 10100001000110010001000010101001010101011111010010 00000100101000000110010100101001000001000000000010 11010000001001110111001001000011101001011011101000 00000110100010001000100000001000011101000000110011 10101000101000100010001111100010101001010000001000 10000010100101001010110000000100101010001011101000 00111100001000010000000110111000000001000000001011 10000001100111010111010001000110111010101101111000
The results can be directly output. (Step 186 results)
输出答案是：DDDDRRURRRRRRDRRRRDDDLDDRDDDDDDDDDDDDRDDRRRURRUURRDDDDRDRRRRRRDRRURRDDDRRRRUURUUUUUUULULLUUUURRRRUULLLUUUULLUUULUURRURRURURRRDDRRRRRDDRRDDLLLDDRRDDRDDLDDDLLDDLLLDLDDDLDDRRRRRRRRRDDDDDDRR
Meaning of the questions:
A lower right corner of FIG output go the shortest path from the particular upper left corner.
Ideas:
Recording a path to the open structure,
Structures in the body need to have x, y coordinates and pre precursors, for recording.
1 #include<stdio.h> 2 #include<iostream> 3 #include<queue> 4 using namespace std; 5 6 struct node 7 { 8 int x,y; 9 } p,q; 10 11 struct nodee 12 { 13 int x,y; 14 char pre; 15 } path[35][55]; 16 17 char a[35][55]; 18 //DLRU 19 //int dir[4][2]= {{0,-1}, {-1,0}, {1,0}, {0,1}}; 20 int dir[4][2]= {{1,0}, {0,-1}, {0,1}, {-1,0}}; 21 22 bool book[35][55]; 23 24 void bfs(int x,int y) 25 { 26 book[x][y]=1; 27 p.x=x; 28 p.y=y; 29 queue<node>Q; 30 Q.push(p); 31 while(!Q.empty()) 32 { 33 q=Q.front(); 34 Q.pop(); 35 for(int i=0; i<4; i++) 36 { 37 p=q; 38 p.x+=dir[i][0]; 39 p.y+=dir[i][1]; 40 if(p.x>=0&&p.x<30&&p.y>=0&&p.y<50&&a[p.x][p.y]=='0'&&!book[p.x][p.y]) 41 { 42 book[p.x][p.y]=1; 43 Q.push(p); 44 path[p.x][p.y].x=q.x; 45 path[p.x][p.y].y=q.y; 46 if(i==0) 47 path[p.x][p.y].pre='D'; 48 if(i==1) 49 path[p.x][p.y].pre='L'; 50 if(i==2) 51 path[p.x][p.y].pre='R'; 52 if(i==3) 53 path[p.x][p.y].pre='U'; 54 if(p.x==29&&p.y==49) 55 return; 56 } 57 } 58 } 59 } 60 61 void w(int x,int y) 62 { 63 if(x==0&&y==0) 64 return; 65 w(path[x][y].x,path[x][y].y); 66 cout<<path[x][y].pre; 67 } 68 69 int main() 70 { 71 for(int i=0; i<30; i++) 72 scanf("%s",a[i]); 73 bfs(0 , 0 ); 74 in ( 29 , 49 ); 75 return 0 ; 76 }