Walk Out

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 141    Accepted Submission(s): 17

Problem Description

In an 

n∗m maze, the right-bottom corner is the exit (position 

(n,m) is the exit). In every position of this maze, there is either a 

0 or a 

1 written on it.

An explorer gets lost in this grid. His position now is 

(1,1), and he wants to go to the exit. Since to arrive at the exit is easy for him, he wants to do something more difficult. At first, he'll write down the number on position 

(1,1). Every time, he could make a move to one adjacent position (two positions are adjacent if and only if they share an edge). While walking, he will write down the number on the position he's on to the end of his number. When finished, he will get a binary number. Please determine the minimum value of this number in binary system.

 

Input

The first line of the input is a single integer 

T (T=10), indicating the number of testcases. 

For each testcase, the first line contains two integers 

n and 

m (1≤n,m≤1000). The 

i-th line of the next 

n lines contains one 01 string of length 

m, which represents 

i-th row of the maze.

 

Output

For each testcase, print the answer in binary system. Please eliminate all the preceding 

0 unless the answer itself is 

0 (in this case, print 

0 instead).

 

Sample Input

2 2 2 11 11 3 3 001 111 101

 

Sample Output

111 101

 

Source

求从1,1到n。n的路径中。得到的二进制数最小是什么

方法：

假设1,1是0那么搜索出全部可达的0号点，前缀0为0，所以能够从这些位置出发。

然后有了一个1,。

那么仅仅要路径最短的，同一时候得到的二进制数最小的。

从全部可出发的点出发，做bfs，计算到n，n的最短路。

然后从n，n做bfs看，那些位置是在最短路上的。

然后从全部可出发的点出发，bfs。

假设存在能到的0点。

则从这些点出发。否则从到的1点出发。

这样贪心做就可以。直到到n。n点

#pragma comment(linker, "/STACK:102400000,102400000")#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           using namespace std;#define maxn 1001char ma[maxn][maxn];int check[maxn][maxn];int dep[maxn][maxn];struct Node{ int x,y; Node(int _x=0,int _y=0):x(_x),y(_y){}};queue
           
             que;vector
            
              ans;vector
             
               one;vector
              
                zero;int n,m;void dfs(int x,int y){ if(check[x][y] || x == n || y == m || x < 0 || y < 0 || ma[x][y] == '1') return; que.push(Node(x,y)); check[x][y] = 1; dfs(x+1,y); dfs(x,y+1); dfs(x-1,y); dfs(x,y-1);}void work(){ while(que.size() > 0) que.pop(); if(ma[0][0] == '1'){ check[0][0] = 1; que.push(Node(0,0)); } else { dfs(0,0); }}void bfs(){ Node a; while(que.size() > 0){ a = que.front(); que.pop(); if(a.x + 1 < n && a.y < m && check[a.x+1][a.y] == 0){ que.push(Node(a.x+1,a.y)); check[a.x+1][a.y] = 1; dep[a.x+1][a.y] = dep[a.x][a.y] + 1; } if(a.x < n && a.y + 1 < m && check[a.x][a.y+1] == 0){ que.push(Node(a.x,a.y+1)); check[a.x][a.y+1] = 1; dep[a.x][a.y+1] = dep[a.x][a.y] + 1; } }}int mark[maxn][maxn];void nidfs(int x,int y){ mark[x][y] = 1; if(x-1>=0 && y >= 0 && mark[x-1][y] == 0 && dep[x-1][y] == dep[x][y] - 1 && dep[x-1][y] > 0) nidfs(x-1,y); if(x >= 0 && y-1>=0 && mark[x][y-1] == 0 && dep[x][y-1] == dep[x][y] - 1 && dep[x][y-1] > 0) nidfs(x,y-1);}void getans(){ ans.clear(); Node a; while(1){ one.clear(); zero.clear(); while(que.size() > 0){ a = que.front(); que.pop(); if(a.x == n -1 && a.y == m -1 ) return ; if(a.x + 1 < n && a.y < m && check[a.x+1][a.y] == 0&& dep[a.x+1][a.y] == dep[a.x][a.y] + 1 && mark[a.x+1][a.y]){ if(ma[a.x+1][a.y] == '1') one.push_back(Node(a.x+1,a.y)); else zero.push_back(Node(a.x+1,a.y)); check[a.x+1][a.y] = 1; } if(a.x < n && a.y + 1< m && check[a.x][a.y+1] == 0 && dep[a.x][a.y+1] == dep[a.x][a.y] + 1 && mark[a.x][a.y+1]){ if(ma[a.x][a.y+1] == '1') one.push_back(Node(a.x,a.y+1)); else zero.push_back(Node(a.x,a.y+1)); check[a.x][a.y+1] = 1; } } if(zero.size() > 0){ ans.push_back(0); for(int i = 0;i < zero.size();i++) que.push(zero[i]); } else { ans.push_back(1); for(int i = 0;i < one.size();i++) que.push(one[i]); } }}int main(){ int t; scanf("%d",&t); while(t--){ scanf("%d%d",&n,&m); for(int i = 0;i < n; i++){ scanf("%s",ma[i]); } memset(check,0,sizeof(check)); work(); memset(dep,0,sizeof(dep)); bfs();// for(int i = 0;i < n; i++){// for(int j = 0;j < m; j++){// printf("%d ",dep[i][j]);// }cout<
               
                < n; i++){// for(int j = 0;j < m; j++){// printf("%d ",dep[i][j]);// }cout<
                
                 < ans.size(); i++) printf("%d",ans[i]); printf("\n"); } } return 0;}/*202 211113 30011111013 31111111113 30000000003 3001011111*/