C. Robot（BFS）

C. Robot

Time Limit: 3000ms

Case Time Limit: 3000ms

Memory Limit: 262144KB

64-bit integer IO format: %lld      Java class name: Main

Submit Status PID:
36000

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There is a rectangular field in the lab of Institution of Advanced Robotic Technology.

Many robots are about to release to this field one after another. They always enter the field from the upper left cell and leave from the lower right cell. In between, they can move vertically or horizontally to an adjacent cell on each step. At any time, there
is at most one robot on the field.

Robots can be divided into two types: Type A and Type B. During the movement, a robot will write down its type mark (A or B) in each cell on its track, which will cover all previous marks in the same cell. Notice that there is no mark on the field at the beginning.

Here is an example: initially there is no mark on the field (Figure 1); first, a robot of type A crosses the field from top-left to bottom right (Figure 2). After that, a robot of type B crosses and its tracks are partially covering the A’s (Figure 3).

                    .....          AAAA.          BBAA.
                    .....          ..AA.          .BBBB
                    .....          ...A.          ...AB
                    .....          .AAA.          .ABBB
                    .....          ..AAA          ..BBB
                     (1)            (2)            (3)

You are given the marks on the field after all robots have crossed. Write a program to determine the minimal possible number of released robots.

Input

The first line contains an integer T (1 ≤ T ≤ 10) -- the number of test cases.



For each test case:

The first line contains two integers h and w, indicates the height and width of the field. 1 ≤ h, w ≤ 4 000.

Then follows h lines, each line contains w characters: each character indicates the mark in the corresponding cell. A dot (“.”) indicates that there is no mark on this cell.

There is at least one mark on the field.

Output

For each test case, output one integer in a single line -- the minimal number of released robots.

Sample Input

2
3 3
AA.
.A.
.AA
5 8
AAB.....
.ABBB...
.AAAAA..
..BBBAAB
.....AAA

Sample Output

1
2

Submit 

cid=5226#status/C" class="button ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" style="margin:0px 0.1em 0px -1px; padding:0px; text-decoration:none; font-family:'Trebuchet MS',Helvetica,Arial,sans-serif; font-size:1.1em; border:1px solid rgb(204,204,204); background-color:rgb(238,238,238); font-weight:bold; color:rgb(68,68,68); display:inline-block; position:relative; zoom:1; overflow:visible">Status 

pid=36000" class="goprob button ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" style="margin:0px 0.1em 0px -1px; padding:0px; text-decoration:none; font-family:'Trebuchet MS',Helvetica,Arial,sans-serif; font-size:1.1em; border:1px solid rgb(204,204,204); background-color:rgb(238,238,238); font-weight:bold; color:rgb(68,68,68); display:inline-block; position:relative; zoom:1; overflow:visible">PID:
36000

#include<stdio.h>
#include<queue>
#include<iostream>
using namespace std;
typedef struct nnn
{
    int x,y;
}node;
int vist[4005][4005];
char map[4005][4005];
int n,m,k,dir[4][2]={0,1,0,-1,1,0,-1,0},tf;

void init()
{
    for(int i=0;i<n;i++)
    for(int j=0;j<m;j++)
    vist[i][j]=0;
}
queue<node>q[2];
void bfs(int flag)
{
    node p,s;
    k++;
    while(!q[flag].empty())
    {
        s=q[flag].front(); q[flag].pop();
        if(s.x==0&&s.y==0)tf=1;
        for(int e=0;e<4;e++)
        {
            p.x=s.x+dir[e][0]; p.y=s.y+dir[e][1];
            if(p.x>=0&&p.x<n&&p.y>=0&&p.y<m&&vist[p.x][p.y]==0&&map[p.x][p.y]!='.')
            {
                vist[p.x][p.y]=1;
                if(map[p.x][p.y]!=map[s.x][s.y])
                    q[!flag].push(p);
                else q[flag].push(p);
            }
        }
    }
}
int main()
{
    int t,f,flag;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)scanf("%s",map[i]);
        k=0; tf=0;
        while(!q[0].empty())q[0].pop();
        while(!q[1].empty())q[1].pop();

        if(map[0][0]!=map[n-1][m-1])f=1;else f=0;
        if(map[n-1][m-1]!='.')
        {
            node s;
            init();
            s.x=n-1; s.y=m-1; flag=0;
            vist[n-1][m-1]=1; q[0].push(s);
           while(!q[flag].empty())
           {
               bfs(flag); flag=(!flag);
           }
        }
        else f=0;
        if(tf==0&&f)f=0;
        printf("%d\n",k-f);
    }
}