HDU (2575) Ocean Currents

Question:
A boat walks in the water and wants to reach its destination. Then there are eight directions. Use 0 ...... 7: North, Northeast, east, southeast ......, But it has a direction at any point, and when its next direction is the same as its own wind direction at the moment (that is, the next direction is the same as its number ), this step does not require energy consumption. Otherwise, the energy consumed is 1 unit of energy. Then the minimum energy required for the boat to reach the destination is required.
Solution:
It seems that the implementation method is relatively easy. The priority queue should be used, but the status mark should be slightly changed. The passing points cannot be simply used with visited [], because one point expands, it is possible that the next point also needs to expand. Because the current point goes to the next point (that is, the point where the previous point expands out) and consumes less energy than the previous one, so we need to use another VST [] to mark the amount of energy that has passed through this point. As long as the energy value reaching this point can be smaller than this point, you can join the team. (Note that each vertex can be taken multiple times. It is important to know the status)

 

# Include <stdio. h>
# Include <string. h>
# Include <queue>
Int dir [8] [2] = },
{1}, {1,-1}, {0,-1}, {-1,-1}; these eight directions are ordered ..
Char map [2100] [2100];
Int VST [2, 2100] [2100];
Using namespace STD;
Int n, m;
Struct point
{
Int X, Y;
Int time;
Friend bool operator <(const point & A, const point & B) // Why does the subsequent Writing Time out? I hope Daniel can explain friend bool operator <(point a, point B)
{
Return A. Time> B. time;
}
};
Int judge (int x, int y)
{
Return x> = 1 & x <= N & Y> = 1 & Y <= m;
}
Int BFS (INT x0, int y0, int X1, int Y1)
{
Int I, j, dx, Dy;
For (I = 1; I <= N; I ++)
For (j = 1; j <= m; j ++)
VST [I] [J] = 999999;
Priority_queue <point> q;
Point In, out;
In. x = x0;
In. Y = y0;
In. Time = 0;
VST [x0] [y0] = 0;
Q. Push (in );
While (! Q. Empty ())
{
Out = Q. Top ();
Q. Pop ();
If (Out. x = x1) & (Out. Y = Y1 ))
Return out. Time;
For (I = 0; I <8; I ++)
{
DX = in. x = out. x + dir [I] [0];
DY = in. Y = out. Y + dir [I] [1];
If (! Judge (dx, Dy ))
Continue;
If (I = (Map [out. x] [out. Y]-'0 '))
In. Time = out. Time;
Else
In. Time = out. Time + 1;
If (in. time <VST [dx] [dy])
{
VST [dx] [dy] = in. time;
Q. Push (in );
}
}
}
Return 0;
}
Int main ()
{
Int I, J, K, h;
Int ex, ey, Sx, Sy;
While (scanf ("% d", & N, & M )! = EOF)
{
For (I = 1; I <= N; I ++)
{
Scanf ("% s", map [I] + 1 );
}
Scanf ("% d", & K );
While (k --)
{
Scanf ("% d", & Ex, & ey, & Sx, & Sy );
Printf ("% d \ n", BFs (ex, ey, Sx, Sy ));
}
}
Return 0;
}