Luo Gu P1141 01 maze BFS

P1141 01 maze

Time limit 1.00s

Memory limit 125.00MB

Title Description

There is only one digital 0 and 1 composed of n- × n-maze grid. If you are in a grid on a 0, then you can move to an adjacent four cells in a cell 1 , the same if you are in a grid 1, then you can move to an adjacent 4 a lattice grid of 0 on the .

Your task is: for a given maze inquiry can begin to move from one cell to the number of grid (including its own).

Input Format

The first 1 Behavior two positive integers n- , m.

Here n lines of n characters, characters can only be 0 or 1, with no spaces between characters.

Subsequently m rows, each row 2 , separated by a space positive integers i, j , corresponding to the first maze i row j a column of the grid, this query initiate movable from cell to cell number.

Output Format

m rows for each query output corresponding answers.

Sample input and output

Input # 1copy

2 2
01
10
1 1
2 2

Output # 1copy

4
4

Description / Tips

All the grid up to each other.

For 2 0 % of the data, n- ≤ . 1 0;

For . 4 0 % of the data, n- ≤ . 5 0;

For . 5 0 % of the data, m ≤ . 5;

For . 6 0 % of the data, n- ≤ . 1 0 0 , m ≤ . 1 0 0;

For . 1 0 0 % of the data, n- ≤ . 1 0 0 0 , m ≤ . 1 0 0 0 0 0.

 

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To thought about this question, we need to pay attention to the following points:

1): for each pass required memory processing, reducing the number of traverse, T is not easy;

2): Do not open unnecessary memset, when a large array, memset command is very time-consuming;

3): when the program runs on a Rockwell valley following the input is \ r \ n two characters, and the program when the unit is tested only following the input \ n a character, it is necessary (after) the original plus a getchar getchar () throw away the rest of the \ n, to prevent the impact on the input map.

 1 #include<stdio.h>
 2 #include<string.h>
 3 int n,m,num,h[3][1000100],u[5]={0,-1,0,1,0},w[5]={0,0,1,0,-1},tap[1010][1010],cu[1000100];
 4 bool map[1010][1010];
 5 int bfs(int,int);
 6 int main(){
 7     memset(tap,0,sizeof(tap));
 8     scanf("%d %d",&n,&m);
 9     getchar();
10     getchar();
11     for(int i=1;i<=n;i++){
12         for(int j=1;j<=n;j++){
13            char o;
14            scanf("%c",&o);
15            if(o=='1') map[i][j]=1;
16            else map[i][j]=0;
17        }
18        getchar();
19        getchar();
20     }
21     num=0;
22     for(int i=1;i<=m;i++){
23         int x,y;
24         scanf("%d %d",&x,&y);
25         if(!tap[x][y]){
26              whether ++ ;
27              Cu [num] = BFS (x, y);
28              printf ( " % d \ n " , since [whether]);
29          }
 30          else printf ( " % d \ n " , since [tap [x] [y]]);
31      }
 32      return  0 ;
33  } 
 34  int BFS ( int sx, int sy) {
 35      int head = 0 , tail = 1 ;
36      h [ 1 ] [tail] = sx; h [2][tail]=sy;
37     tap[sx][sy]=num;
38     do{
39         head++;
40         for(int i=1;i<=4;i++){
41             int x,y;
42             x=h[1][head]+u[i];
43             y=h[2][head]+w[i];
44             if(x>0&&y>0&&x<=n&&y<=n&&(!tap[x][y]))
45                if((map[h[1][head]][h[2][head]]&&!map[x][y])||(!map[h[1][head]][h[2][head]]&&map[x][y])){
46                    tail++;
47                    h[1][tail]=x;
48                    h[2][tail]=y;
49                    tap[x][y]=num;
50                }
51         }
52     }while(head<tail);
53     return tail;
54 }

BFS