Borg Maze poj3026

First of all, this question has to let me speak out. The test data is too speechless. After the input columns and rows, spaces will still be entered, so getchar () cannot be used () to process '\ n', only gets () can be used, so that space and' \ n' can be processed at one time '.

Then let's talk about the idea of this question. I first use BFS for each A or S to find the shortest distance between it and other A or S, then, create A graph with these A or S, calculate the minimum spanning tree, and then add all edge weights.

 

[Cpp]
# Include <iostream>
# Include <cstdio>
# Include <cstring>
Using namespace std;
Struct st
{
Int x, y, step;
} W, tmp;
St q [2, 2505];
Constint INF = 100000000;
Int map [55] [55];
Bool visit [55] [55];
Bool visit1 [105];
Int dist [105] [105];
Int dirt [4] [2] = {, 0,-, 0 };
Int d [105];
Int t, n, m, num;
 
Bool OK (int x, int y)
{
Return x> = 0 & x <n & y> = 0 & y <m & map [x] [y]! =-1;
}
 
Void BFS (int u)
{
Memset (visit, false, sizeof (visit ));
Visit [w. x] [w. y] = true;
Int front = 0, rear = 0;
Q [rear ++] = w;
While (front <rear)
{
W = q [front ++];
If (map [w. x] [w. y]> 0)
Dist [u] [map [w. x] [w. y] = w. step;
For (int I = 0; I <4; I ++)
{
Tmp. x = w. x + dirt [I] [0];
Tmp. y = w. y + dirt [I] [1];
Tmp. step = w. step + 1;
If (OK (tmp. x, tmp. y )&&! Visit [tmp. x] [tmp. y])
{
Visit [tmp. x] [tmp. y] = true;
Q [rear ++] = tmp;
}
}
}
}
 
Int main ()
{
Int I, j, k, sum;
Scanf ("% d", & t );
Char tmp [100];
While (t --)
{
Num = 0;
Scanf ("% d", & m, & n );
// Getchar ();
Gets (tmp );
For (I = 0; I <n; I ++)
{
Gets (tmp );
For (j = 0; j <m; j ++)
{
If (tmp [j] = 's' | tmp [j] = 'A ')
Map [I] [j] = ++ num;
Else if (tmp [j] = '')
Map [I] [j] = 0;
Else
Map [I] [j] =-1;
}
}
// Calculate the shortest distance of any two alien instances.
For (I = 0; I <n; I ++)
For (j = 0; j <m; j ++)
If (map [I] [j]> 0)
{
W. x = I; w. y = j; w. step = 0;
BFS (map [I] [j]);
}
 
// Prim
For (I = 0; I <= num; I ++)
{
Visit1 [I] = false;
D [I] = INF;
}
D [1] = 0;
For (I = 1; I <= num; I ++)
{
K = 0;
For (j = 1; j <= num; j ++)
If (d [j] <d [k] &! Visit1 [j])
K = j;
If (k = 0)
Break;
Visit1 [k] = true;
For (j = 1; j <= num; j ++)
If (! Visit1 [j] & dist [k] [j] <d [j])
D [j] = dist [k] [j];
}
Sum = 0;
For (j = 1; j <= num; j ++)
Sum + = d [j];
Printf ("% d \ n", sum );
}
Return 0;
}