HDU 5040 BFS + pressure, hdu5040bfs

HDU 5040 BFS + pressure, hdu5040bfs

2014 ACM/ICPC Asia Regional Beijing Online

For N * N Matrices

M start point, T end point

Cameras with starting directions north N, East E, South S, and West W can be detected in the range of self-directed 1 grid and Rotate 90 ° clockwise every 1 second.

If there is a lamp in front or a place on your own, it takes 3 seconds to move it, or wait for a second to move it.

BFS priority queue

Enable the three-dimensional hash array to determine the weight. The third dimension is the waiting time at this point. Open it to 4 (the camera turns around)

When will the camera view the processing of each vertex in the figure, and store it in the MAP array with a binary compression?

Then the regular BFS Solution

In the competition, no CASE was left blank .... I found a half-day error .....

/* _ OoOoo _ o8888888o 88 ". "88 (|-_-|) O \=/O ____/'---'\____. '\\| | //'. /\ |: | // \/_ | -: -|-\ | \-// | \ _ | ''\ ---/'' | \. -\__'-'___/-. /___'.. '/--. --\'.. __. "" '<'. ___ \_< |> _/___. '> '"". | :'-\'.; '\_/';. '/-': | \\'-. \___\/___/. -'// ====== '-. ____'-. ___\_____/___. -'____. -'= --- =' ^ ^ fozu Baoyou pass System Test! */# Include "stdio. h "# include" string. h "# include" queue "using namespace std; int inf = 0x3f; int dir [4] [2] = }, {0,-1 }}; struct node {int x, y, step, wait; // step records the number of steps and wait records the waiting time of friend bool operator <(node n1, node n2) {return n1.step> n2.step; }}; int s_x, s_y, n; int B [5]; char str [510] [510]; int map [510] [510], hash [510] [510] [5]; // when will each point of MAP storage be seen, HASH weight the third dimension is the waiting time at this point int bfs () {priority_queue <no De> q; node cur, next; int I, time; cur. x = s_x; cur. y = s_y; cur. step = 0; cur. wait = 0; q. push (cur); memset (hash, inf, sizeof (hash); hash [cur. x] [cur. y] [0] = 0; while (! Q. empty () {cur = q. top (); q. pop (); if (str [cur. x] [cur. y] = 'T') return cur. step; for (I = 0; I <4; I ++) {next. x = cur. x + dir [I] [0]; next. y = cur. y + dir [I] [1]; if (next. x <0 | next. x> = n | next. y <0 | next. y> = n) continue; if (str [next. x] [next. y] = '#') continue; time = B [cur. step % 4]; if (map [cur. x] [cur. y] & time) = time | (map [next. x] [next. y] & time) = time) // next is displayed at the current or next vertex. step = cur. step + 3; else next. step = cur. step + 1; next. wait = 0; if (next. step 










Why is hdu 1181 bfs wa?
/*
A set of data for the landlord
Bb
Tm
0
======================================
There is a small bug, and the AC will be changed.
*/
# Include <iostream>
# Include <string. h>
# Include <queue>
Using namespace std;
# Define N 10020
Struct Point
{
Char pre, end;
};
Point map [N];
Bool vis [N];
Int n;

Void BFS ()
{
Int I;
Char;
Queue <char> Q;
Memset (vis, false, sizeof (vis ));
For (I = 0; I <n; I ++)
If (map [I]. pre = 'B ')
{
Q. push (map [I]. end );
}
 
While (! Q. empty ())
{
A = Q. front ();
Q. pop ();
For (I = 0; I <n; I ++)
{
If (map [I]. pre = a & map [I]. end = 'M ') //!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
{
Printf ("Yes. \ n ");
Return;
}
If (! Vis [I] & map [I]. pre =)
{
Vis [I] = true;
Q. push (map [I]. end );
}
}
}
Printf ("No. \ n ");
}

Int main ()
{
Char str [102];
N = 0;
While (scanf ("% s", str )! = EOF)
{
If (str [0] = '0 ')
{
BFS ();
N = 0;
Continue;
}
Map [n]. pre = str [0];
Map [n]. end = str [(int) strlen (str)-1];
N ++;
}
Return 0;
}... Remaining full text>

Does hdu hdoj 1010 support BFS?
Yes .. Queue struct, a coordinate, a time ..
YES when the time is equal and the coordinates are equal.
When the queue header time is greater than T, NO.
I did not know whether it is feasible ..
0 Ms code ..
# Include <stdio. h>
# Include <string. h>
Const int dir [4] [2] = {}, {0,-1 }};
Const int MAX = 8;
Int n, m;
Char map [MAX] [MAX];
Int dx, dy;
Struct
{
Int x, y;
} Q [2, 1000000];
Int used [MAX] [MAX];
Int vabs (int x)
{
Return x <0? -X: x;
}
Int dis (int x, int y)
{
Return vabs (x) + vabs (y );
}
Int BFS (int x, int y) // first checks whether a path exists.
{
Int tx, ty, I, f =-1, r = 0;
Memset (used, 0, sizeof (used ));
Used [x] [y] = 1;
Q [0]. x = x;
Q [0]. y = y;
While (f! = R ){
F ++;
X = q [f]. x;
Y = q [f]. y;
For (I = 0; I <4; I ++ ){
Tx = x + dir [I] [0];
Ty = y + dir [I] [1];
If (tx = dx & ty = dy) return 1;
If (tx <0 | tx> = n | ty <0 | ty> = m | used [tx] [ty]) continue;
If (map [tx] [ty] = 'X') continue;
Used [tx] [ty] = 1; // mark the point
++ R;
Q [r]. x = tx;
Q [r]. y = ty;
}
}
Return 0;
}
Int able (int x, int y)
{
Memset (used, 0, sizeof (used ));
Int I, f =-1, r = 0, tx, ty, cnt = 0;
Q [0]. x = x;
Q [0]. y = y;
Used [x] [y] = 1;
While (f! = R ){
F ++;
X = q [f]. x;
Y = q [f]. y;
For (I = 0; I <4; I ++ ){
Tx = dir [I] [0] + x;
Ty = dir [I] [1] + y;
If (tx <0 | tx> = n | ty <0 | ty> = m | used [tx] [ty]) continue;
If (map [tx] [ty] = 'X') continue;
Cnt ++;
Used [tx] [ty] = 1;
R ++;
Q [r... the remaining full text>