/*
* Sorting hook srt squares by "distance to player"
*
* We use pythagoreans theorum to calculate the distance
*/
static bool coords_sort_distance(coord c1, coord c2)
{
int py = p_ptr->py;
int px = p_ptr->px;
int dist1 = GET_SQUARE(py - c1.y) + GET_SQUARE(px - c1.x);
int dist2 = GET_SQUARE(py - c2.y) + GET_SQUARE(px - c2.x);
/* Compare the distances */
return (dist1 <= dist2);
}
static void game_redraw(drawing *dr, game_drawstate *ds,
const game_state *oldstate, const game_state *state,
int dir, const game_ui *ui,
float animtime, float flashtime)
{
int i, j;
struct bbox bb = find_bbox(&state->params);
struct solid *poly;
const int *pkey, *gkey;
float t[3];
float angle;
int square;
draw_rect(dr, 0, 0, XSIZE(GRID_SCALE, bb, state->solid),
YSIZE(GRID_SCALE, bb, state->solid), COL_BACKGROUND);
if (dir < 0) {
const game_state *t;
/*
* This is an Undo. So reverse the order of the states, and
* run the roll timer backwards.
*/
assert(oldstate);
t = oldstate;
oldstate = state;
state = t;
animtime = ROLLTIME - animtime;
}
if (!oldstate) {
oldstate = state;
angle = 0.0;
square = state->current;
pkey = state->dpkey;
gkey = state->dgkey;
} else {
angle = state->angle * animtime / ROLLTIME;
square = state->previous;
pkey = state->spkey;
gkey = state->sgkey;
}
state = oldstate;
for (i = 0; i < state->grid->nsquares; i++) {
int coords[8];
for (j = 0; j < state->grid->squares[i].npoints; j++) {
coords[2*j] = ((int)(state->grid->squares[i].points[2*j] * GRID_SCALE)
+ ds->ox);
coords[2*j+1] = ((int)(state->grid->squares[i].points[2*j+1]*GRID_SCALE)
+ ds->oy);
}
draw_polygon(dr, coords, state->grid->squares[i].npoints,
GET_SQUARE(state, i) ? COL_BLUE : COL_BACKGROUND,
COL_BORDER);
}
/*
* Now compute and draw the polyhedron.
*/
poly = transform_poly(state->solid, state->grid->squares[square].flip,
pkey[0], pkey[1], angle);
/*
* Compute the translation required to align the two key points
* on the polyhedron with the same key points on the current
* face.
*/
for (i = 0; i < 3; i++) {
float tc = 0.0;
for (j = 0; j < 2; j++) {
float grid_coord;
if (i < 2) {
grid_coord =
state->grid->squares[square].points[gkey[j]*2+i];
} else {
grid_coord = 0.0;
}
tc += (grid_coord - poly->vertices[pkey[j]*3+i]);
}
t[i] = tc / 2;
}
for (i = 0; i < poly->nvertices; i++)
for (j = 0; j < 3; j++)
poly->vertices[i*3+j] += t[j];
/*
* Now actually draw each face.
*/
for (i = 0; i < poly->nfaces; i++) {
float points[8];
int coords[8];
for (j = 0; j < poly->order; j++) {
int f = poly->faces[i*poly->order + j];
points[j*2] = (poly->vertices[f*3+0] -
poly->vertices[f*3+2] * poly->shear);
points[j*2+1] = (poly->vertices[f*3+1] -
poly->vertices[f*3+2] * poly->shear);
}
for (j = 0; j < poly->order; j++) {
coords[j*2] = (int)floor(points[j*2] * GRID_SCALE) + ds->ox;
coords[j*2+1] = (int)floor(points[j*2+1] * GRID_SCALE) + ds->oy;
}
/*
* Find out whether these points are in a clockwise or
* anticlockwise arrangement. If the latter, discard the
* face because it's facing away from the viewer.
*
* This would involve fiddly winding-number stuff for a
* general polygon, but for the simple parallelograms we'll
* be seeing here, all we have to do is check whether the
* corners turn right or left. So we'll take the vector
* from point 0 to point 1, turn it right 90 degrees,
* and check the sign of the dot product with that and the
* next vector (point 1 to point 2).
*/
{
float v1x = points[2]-points[0];
float v1y = points[3]-points[1];
float v2x = points[4]-points[2];
float v2y = points[5]-points[3];
float dp = v1x * v2y - v1y * v2x;
if (dp <= 0)
continue;
}
draw_polygon(dr, coords, poly->order,
state->facecolours[i] ? COL_BLUE : COL_BACKGROUND,
COL_BORDER);
}
sfree(poly);
draw_update(dr, 0, 0, XSIZE(GRID_SCALE, bb, state->solid),
YSIZE(GRID_SCALE, bb, state->solid));
/*
* Update the status bar.
*/
{
char statusbuf[256];
if (state->completed) {
strcpy(statusbuf, _("COMPLETED!"));
strcpy(statusbuf+strlen(statusbuf), " ");
} else statusbuf[0] = '\0';
sprintf(statusbuf+strlen(statusbuf), _("Moves: %d"),
(state->completed ? state->completed : state->movecount));
status_bar(dr, statusbuf);
}
}
static game_state *execute_move(const game_state *from, const char *move)
{
game_state *ret;
float angle;
struct solid *poly;
int pkey[2];
int skey[2], dkey[2];
int i, j, dest;
int direction;
switch (*move) {
case 'L': direction = LEFT; break;
case 'R': direction = RIGHT; break;
case 'U': direction = UP; break;
case 'D': direction = DOWN; break;
default: return NULL;
}
dest = find_move_dest(from, direction, skey, dkey);
if (dest < 0)
return NULL;
ret = dup_game(from);
ret->current = dest;
/*
* So we know what grid square we're aiming for, and we also
* know the two key points (as indices in both the source and
* destination grid squares) which are invariant between source
* and destination.
*
* Next we must roll the polyhedron on to that square. So we
* find the indices of the key points within the polyhedron's
* vertex array, then use those in a call to transform_poly,
* and align the result on the new grid square.
*/
{
int all_pkey[4];
align_poly(from->solid, &from->grid->squares[from->current], all_pkey);
pkey[0] = all_pkey[skey[0]];
pkey[1] = all_pkey[skey[1]];
/*
* Now pkey[0] corresponds to skey[0] and dkey[0], and
* likewise [1].
*/
}
/*
* Now find the angle through which to rotate the polyhedron.
* Do this by finding the two faces that share the two vertices
* we've found, and taking the dot product of their normals.
*/
{
int f[2], nf = 0;
float dp;
for (i = 0; i < from->solid->nfaces; i++) {
int match = 0;
for (j = 0; j < from->solid->order; j++)
if (from->solid->faces[i*from->solid->order + j] == pkey[0] ||
from->solid->faces[i*from->solid->order + j] == pkey[1])
match++;
if (match == 2) {
assert(nf < 2);
f[nf++] = i;
}
}
assert(nf == 2);
dp = 0;
for (i = 0; i < 3; i++)
dp += (from->solid->normals[f[0]*3+i] *
from->solid->normals[f[1]*3+i]);
angle = (float)acos(dp);
}
/*
* Now transform the polyhedron. We aren't entirely sure
* whether we need to rotate through angle or -angle, and the
* simplest way round this is to try both and see which one
* aligns successfully!
*
* Unfortunately, _both_ will align successfully if this is a
* cube, which won't tell us anything much. So for that
* particular case, I resort to gross hackery: I simply negate
* the angle before trying the alignment, depending on the
* direction. Which directions work which way is determined by
* pure trial and error. I said it was gross :-/
*/
{
int all_pkey[4];
int success;
if (from->solid->order == 4 && direction == UP)
angle = -angle; /* HACK */
poly = transform_poly(from->solid,
from->grid->squares[from->current].flip,
pkey[0], pkey[1], angle);
flip_poly(poly, from->grid->squares[ret->current].flip);
success = align_poly(poly, &from->grid->squares[ret->current], all_pkey);
if (!success) {
sfree(poly);
angle = -angle;
poly = transform_poly(from->solid,
from->grid->squares[from->current].flip,
pkey[0], pkey[1], angle);
flip_poly(poly, from->grid->squares[ret->current].flip);
success = align_poly(poly, &from->grid->squares[ret->current], all_pkey);
}
assert(success);
}
/*
* Now we have our rotated polyhedron, which we expect to be
* exactly congruent to the one we started with - but with the
* faces permuted. So we map that congruence and thereby figure
* out how to permute the faces as a result of the polyhedron
* having rolled.
*/
{
int *newcolours = snewn(from->solid->nfaces, int);
for (i = 0; i < from->solid->nfaces; i++)
newcolours[i] = -1;
for (i = 0; i < from->solid->nfaces; i++) {
int nmatch = 0;
/*
* Now go through the transformed polyhedron's faces
* and figure out which one's normal is approximately
* equal to this one.
*/
for (j = 0; j < poly->nfaces; j++) {
float dist;
int k;
dist = 0;
for (k = 0; k < 3; k++)
dist += SQ(poly->normals[j*3+k] -
from->solid->normals[i*3+k]);
if (APPROXEQ(dist, 0)) {
nmatch++;
newcolours[i] = ret->facecolours[j];
}
}
assert(nmatch == 1);
}
for (i = 0; i < from->solid->nfaces; i++)
assert(newcolours[i] != -1);
sfree(ret->facecolours);
ret->facecolours = newcolours;
}
ret->movecount++;
/*
* And finally, swap the colour between the bottom face of the
* polyhedron and the face we've just landed on.
*
* We don't do this if the game is already complete, since we
* allow the user to roll the fully blue polyhedron around the
* grid as a feeble reward.
*/
if (!ret->completed) {
i = lowest_face(from->solid);
j = ret->facecolours[i];
ret->facecolours[i] = GET_SQUARE(ret, ret->current);
SET_SQUARE(ret, ret->current, j);
/*
* Detect game completion.
*/
j = 0;
for (i = 0; i < ret->solid->nfaces; i++)
if (ret->facecolours[i])
j++;
if (j == ret->solid->nfaces) {
ret->completed = ret->movecount;
}
}
sfree(poly);
/*
* Align the normal polyhedron with its grid square, to get key
* points for non-animated display.
*/
{
int pkey[4];
int success;
success = align_poly(ret->solid, &ret->grid->squares[ret->current], pkey);
assert(success);
ret->dpkey[0] = pkey[0];
ret->dpkey[1] = pkey[1];
ret->dgkey[0] = 0;
ret->dgkey[1] = 1;
}
ret->spkey[0] = pkey[0];
ret->spkey[1] = pkey[1];
ret->sgkey[0] = skey[0];
ret->sgkey[1] = skey[1];
ret->previous = from->current;
ret->angle = angle;
return ret;
}
#define _GNU_SOURCE
#include <stdlib.h>
#include <stdarg.h>
#include <stdio.h>
#include <string.h>
#include "monitor.h"
#include "serveur.h"
void avertCase(serveur* this, clients* monitor, char* k) {
position p;
char* r;
sscanf(k, "bct %d %d\n", &(p.x), &(p.y));
r = GET_SQUARE(this, p.x, p.y);
avertThisMonitor(monitor,
mCaseMap(p.x, p.y,
r[nourriture], r[linemate], r[deraumere],
r[sibur], r[mendiane], r[phiras],
r[thystame]
));
}
void avertMap(serveur* this, clients* monitor) {
position p;
char* r;
p.x = 0;
while (p.x < this->size.x)
{
p.y = 0;
