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Fix loophole in Palisade completion checker.

A user pointed out that if you construct a 'solution' in which no clue
square has too _many_ borders but at least one has too few, and then
bring those clues up to their count by adding extra stray border lines
_inside_ a connected component (avoiding actually dividing any
component completely into two), then the game checker treats that as
solved for victory-flash purposes, on the grounds that (a) the grid is
divided into components of the right size and (b) all clues are
satisfied.

A small example is 4x4n4:22a2b2c33, with the non-solution of dividing
the grid into four 2x2 square blocks and then adding a spurious extra
edge between the two 3 clues. The old Palisade completion check would
flash for victory _at the same time_ as highlighting the spurious edge
in COL_ERROR.

Fixed by enforcing in is_solved() that every border line must separate
two distinct connected components.

--- a/palisade.c

+++ b/palisade.c

@@ -511,15 +511,45 @@

 static int is_solved(const game_params *params, clue *clues,

                      borderflag *border)

 {

-    int wh = params->w * params->h, k = params->k, *dsf = snew_dsf(wh), i;

+    int w = params->w, h = params->h, wh = w*h, k = params->k;

+    int i, x, y;

+    int *dsf = snew_dsf(wh);

     assert (dsf[0] == UNVISITED); /* check: UNVISITED and dsf.c match up */

+    /*

+     * A game is solved if:

+     *

+     *  - the borders drawn on the grid divide it into connected

+     *    components such that every square is in a component of the

+     *    correct size

+     *  - the borders also satisfy the clue set

+     */

     for (i = 0; i < wh; ++i) {

         if (dsf[i] == UNVISITED) dfs_dsf(i, params->w, border, dsf, TRUE);

         if (dsf_size(dsf, i) != k) goto error;

         if (clues[i] == EMPTY) continue;

         if (clues[i] != bitcount[border[i] & BORDER_MASK]) goto error;

+    }

+

+    /*

+     * ... and thirdly:

+     *

+     *  - there are no *stray* borders, in that every border is

+     *    actually part of the division between two components.

+     *    Otherwise you could cheat by finding a subdivision which did

+     *    not *exceed* any clue square's counter, and then adding a

+     *    few extra edges.

+     */

+    for (y = 0; y < h; y++) {

+        for (x = 0; x < w; x++) {

+            if (x+1 < w && (border[y*w+x] & BORDER_R) &&

+                dsf_canonify(dsf, y*w+x) == dsf_canonify(dsf, y*w+(x+1)))

+                goto error;

+            if (y+1 < h && (border[y*w+x] & BORDER_D) &&

+                dsf_canonify(dsf, y*w+x) == dsf_canonify(dsf, (y+1)*w+x))

+                goto error;

+        }

     }

     sfree(dsf);