Problem Description


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


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


0 unless the answer itself is 0 (in this case, print 0


 



Sample Input


2 2 2 11 11 3 3 001 111 101


 



Sample Output


101


先用dfs判断最远的0可以到达的位置，然后按照斜行递推

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<algorithm>
#include<string>
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
using namespace std;
const ll maxn=1005;
int T,n,m,f[maxn][maxn],c[maxn][maxn],tot;
char s[maxn][maxn];

void dfs(int x,int y)
{
    if (c[x][y]) return ;
    c[x][y]=1;
    if (s[x][y]=='1') return;
    f[x][y]=1;
    if(x+y>tot) tot=x+y;
    if (x>1) dfs(x-1,y);
    if (x<n) dfs(x+1,y);
    if (y>1) dfs(x,y-1);
    if (y<m) dfs(x,y+1);
}

int main()
{
    scanf("%d",&T);
    while (T--)
    {
        memset(f,0,sizeof(f));
        memset(c,0,sizeof(c));
        scanf("%d%d",&n,&m);
        tot=1;
        for (int i=1;i<=n;i++)
        {
            scanf("%s",s[i]+1);
        }
        dfs(1,1);
        if(tot==n+m)
        {printf("0\n");continue;}
        if(tot==1)
        {
            tot=2;
            f[1][1]=1;
            printf("%d",1);
        }
        for (int i=tot;i<n+m;i++)
        {
            int flag=1;
            for (int j=max(1,i-m);j<=min(n,i-1);j++)
            if (f[j][i-j])
            {
                int x=j<n?s[j+1][i-j]-'0':1;
                int y=i-j<m?s[j][i-j+1]-'0':1;
                flag=min(flag,min(x,y));
            }
            for (int j=max(1,i-m);j<=min(n,i-1);j++)
            if (f[j][i-j])
            {
                int x=j<n?s[j+1][i-j]-'0':1;
                int y=i-j<m?s[j][i-j+1]-'0':1;
                if (x==flag) f[j+1][i-j]=1;
                if (y==flag) f[j][i-j+1]=1;
            }
            printf("%d",flag);
        }
        printf("\n");
    }
    return 0;
}