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FruitFrolic
这是一个连连看小游戏，以 Unity2D 开发。因用了数种水果图片来做头像，所以游戏取名 FruitFrolic。同样，它也只是我闲时的练手。
少时曾玩过掌上游戏机里的俄罗斯方块及打飞机，及手机上的推箱子等，也在 Dos 上玩过几乎人人皆知的超级玛丽。我很想在闲暇的时候自己来实现它们，但为兴趣和乐趣而已。所以有前文所述的 PetGenie，以及本文，和之后可能的自实现版俄罗斯方块。不过限于美术素材及个人精力等之因，它们应会实现得比较简陋，虽然游戏核心逻辑几都具备。
而我所使用的所有美术素材及音频等都来源于网络，本着开放的原则，我的（所有）自实现小游戏也都开源，且没有任何版权等限制。
连连看的核心显然在洗牌及连线分析算法。洗牌控制了游戏的难易，变化很多。但我这里只是简单地平均生成了头像并随机打乱，而在连线分析算法里使用了广度优先搜索（参考《编程之美》的论述以代码实现）。
洗牌代码如下。
void RandomGenies() { int idx; // 共 6 * 8 个格子、12 种水果 --> 每种水果生成 4 次 int[] geniesCounter = new int[cSpriteTypeCount] {4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4}; for (int i = 0; i < cGridRows; i++) { for (int j = 0; j < cGridCols; j++) { genies[i, j].x = i; genies[i, j].y = j; while (true) { idx = (int)(Random.value * cSpriteTypeCount); if (geniesCounter[idx] > 0) { break; } } geniesCounter[idx]--; genies[i, j].index = idx; genies[i, j].spriteRenderer.sprite = fruitSprites[idx]; } } }
而连线分析实现代码如下。
void DetectLink() { if (!ValidPrecondition()) { return; } List<TGenie> contnr0 = new List<TGenie>(); FindCells((int)(touchCoords.pos1.x), (int)(touchCoords.pos1.y), contnr0); if (CellExists(genies[(int)(touchCoords.pos2.y), (int)(touchCoords.pos2.x)], contnr0)) { genies[(int)(touchCoords.pos1.y), (int)(touchCoords.pos1.x)].index = cInvalidCoordValue; genies[(int)(touchCoords.pos2.y), (int)(touchCoords.pos2.x)].index = cInvalidCoordValue; return; } List<TGenie> contnr1 = new List<TGenie>(); PrepareContnr(contnr0, contnr1); ShrinkContnr(contnr0, contnr1); if (CellExists(genies[(int)(touchCoords.pos2.y), (int)(touchCoords.pos2.x)], contnr1)) { genies[(int)(touchCoords.pos1.y), (int)(touchCoords.pos1.x)].index = cInvalidCoordValue; genies[(int)(touchCoords.pos2.y), (int)(touchCoords.pos2.x)].index = cInvalidCoordValue; return; } List<TGenie> contnr2 = new List<TGenie>(); PrepareContnr(contnr1, contnr2); ShrinkContnr(contnr0, contnr2); ShrinkContnr(contnr1, contnr2); if (CellExists(genies[(int)(touchCoords.pos2.y), (int)(touchCoords.pos2.x)], contnr2)) { genies[(int)(touchCoords.pos1.y), (int)(touchCoords.pos1.x)].index = cInvalidCoordValue; genies[(int)(touchCoords.pos2.y), (int)(touchCoords.pos2.x)].index = cInvalidCoordValue; return; } DetectSpecialLink(); }bool ValidPrecondition() { if ((touchCoords.pos1.x == cInvalidCoordValue) || (touchCoords.pos1.y == cInvalidCoordValue) || (touchCoords.pos2.x == cInvalidCoordValue) || (touchCoords.pos2.y == cInvalidCoordValue)) { return false; } if ((touchCoords.pos1.x == touchCoords.pos2.x) && (touchCoords.pos1.y == touchCoords.pos2.y)) { return false; } if (genies[(int)(touchCoords.pos1.y), (int)(touchCoords.pos1.x)].index != genies[(int)(touchCoords.pos2.y), (int)(touchCoords.pos2.x)].index) { return false; } return true; }void FindCells(int x, int y, List<TGenie> contnr) { int n = 0; n = y - 1; if (n >= 0) { if (!(contnr.Contains(genies[n, x]))) { contnr.Add(genies[n, x]); } } while ((n >= 0) && (genies[n, x].index == cInvalidCoordValue)) { n--; if ((n >= 0) && (!(contnr.Contains(genies[n, x])))) { contnr.Add(genies[n, x]); } } n = y + 1; if (n < cGridRows) { if (!(contnr.Contains(genies[n, x]))) { contnr.Add(genies[n, x]); } } while ((n < cGridRows) && (genies[n, x].index == cInvalidCoordValue)) { n++; if ((n < cGridRows) && (!(contnr.Contains(genies[n, x])))) { contnr.Add(genies[n, x]); } } n = x - 1; if (n >= 0) { if (!(contnr.Contains(genies[y, n]))) { contnr.Add(genies[y, n]); } } while ((n >= 0) && (genies[y, n].index == cInvalidCoordValue)) { n--; if ((n >= 0) && (!(contnr.Contains(genies[y, n])))) { contnr.Add(genies[y, n]); } } n = x + 1; if (n < cGridCols) { if (!(contnr.Contains(genies[y, n]))) { contnr.Add(genies[y, n]); } } while ((n < cGridCols) && (genies[y, n].index == cInvalidCoordValue)) { n++; if ((n < cGridCols) && (!(contnr.Contains(genies[y, n])))) { contnr.Add(genies[y, n]); } } } bool CellExists(TGenie genie, List<TGenie> contnr) { foreach (TGenie g in contnr) { if (g.Equals(genie)) { return true; } } return false; } void PrepareContnr(List<TGenie> contnrSrc, List<TGenie> contnrDest) { foreach (TGenie g in contnrSrc) { if (g.index == cInvalidCoordValue) { FindCells(g.y, g.x, contnrDest); } } } void ShrinkContnr(List<TGenie> contnrSrc, List<TGenie> contnrDest) { foreach (TGenie g in contnrSrc) { if (contnrDest.Contains(g)) { contnrDest.Remove(g); } } }void DetectSpecialLink() { // 若在第一或最末列 if (touchCoords.pos1.x == touchCoords.pos2.x) { if ((touchCoords.pos1.x == 0) || (touchCoords.pos1.x == cGridCols - 1)) { genies[(int)(touchCoords.pos1.y), (int)(touchCoords.pos1.x)].index = cInvalidCoordValue; genies[(int)(touchCoords.pos2.y), (int)(touchCoords.pos2.x)].index = cInvalidCoordValue; return; } } // 若在第一或最末行 if (touchCoords.pos1.y == touchCoords.pos2.y) { if ((touchCoords.pos1.y == 0) || (touchCoords.pos1.y == cGridRows - 1)) { genies[(int)(touchCoords.pos1.y), (int)(touchCoords.pos1.x)].index = cInvalidCoordValue; genies[(int)(touchCoords.pos2.y), (int)(touchCoords.pos2.x)].index = cInvalidCoordValue; return; } } }
判断是否已连线完毕（已全部消除或已死锁）的代码如下。
bool HasMatches() { // 检测上下左右第一行/列是否有可消除的格子(特殊处理) for (int i = 0; i < cGridCols - 1; i++) { if (genies[0, i].index != cInvalidCoordValue) { for (int j = i + 1; j < cGridCols; j++) { if (genies[0, i].index == genies[0, j].index) { return true; } } } if (genies[cGridRows - 1, i].index != cInvalidCoordValue) { for (int j = i + 1; j < cGridCols; j++) { if (genies[cGridRows - 1, i].index == genies[cGridRows - 1, j].index) { return true; } } } } for (int i = 0; i < cGridRows - 1; i++) { if (genies[i, 0].index != cInvalidCoordValue) { for (int j = i + 1; j < cGridRows; j++) { if (genies[i, 0].index == genies[j, 0].index) { return true; } } } if (genies[i, cGridCols - 1].index != cInvalidCoordValue) { for (int j = i + 1; j < cGridRows; j++) { if (genies[i, cGridCols - 1].index == genies[j, cGridCols - 1].index) { return true; } } } } for (int i = 0; i < cGridRows; i++) { for (int j = 0; j < cGridCols; j++) { if (genies[i, j].index != cInvalidCoordValue) { // 0 转弯 List<TGenie> contnr0 = new List<TGenie>(); FindCells(j, i, contnr0); if (HasMatchableGenie(genies[i, j], contnr0)) { return true; } // 1 转弯 List<TGenie> contnr1 = new List<TGenie>(); PrepareContnr(contnr0, contnr1); ShrinkContnr(contnr0, contnr1); if (HasMatchableGenie(genies[i, j], contnr1)) { return true; } // 2 转弯 List<TGenie> contnr2 = new List<TGenie>(); PrepareContnr(contnr1, contnr2); ShrinkContnr(contnr0, contnr2); ShrinkContnr(contnr1, contnr2); RemoveNullGenies(contnr2); if (HasMatchableGenie(genies[i, j], contnr2)) { return true; } } } } return false; }bool HasMatchableGenie(TGenie genie, List<TGenie> contnr) { foreach (TGenie g in contnr) { if ((!g.Equals(genie)) && (g.index == genie.index)) { return true; } } return false; } void RemoveNullGenies(List<TGenie> contnr) { List<TGenie> tmp = new List<TGenie>(); foreach (TGenie g in contnr) { if (g.index == cInvalidCoordValue) { tmp.Add(g); } } foreach (TGenie g in tmp) { contnr.Remove(g); } }
其实我本想分析每一种可能的连线情况（0---2 个转弯），但在写完 0 和 1 个转弯分析之后不想再写 2 个转弯分析代码了，因它们确实不好理解（也不好维护）。
// 0 个转角连通 bool CheckLink0() { // 若在同一列格子 if (touchCoords.pos1.x == touchCoords.pos2.x) { if ((touchCoords.pos1.x != 0) && (touchCoords.pos1.x != cGridCols - 1)) { if (touchCoords.pos1.y < touchCoords.pos2.y) { for (int i = (int)(touchCoords.pos1.y) + 1; i < ((int)(touchCoords.pos2.y)); i++) { if (genies[i, (int)(touchCoords.pos1.x)].index != cInvalidCoordValue) { return false; } } } else { for (int i = (int)(touchCoords.pos2.y) + 1; i < ((int)(touchCoords.pos1.y)); i++) { if (genies[i, (int)(touchCoords.pos1.x)].index != cInvalidCoordValue) { return false; } } } } genies[(int)(touchCoords.pos1.y), (int)(touchCoords.pos1.x)].index = cInvalidCoordValue; genies[(int)(touchCoords.pos2.y), (int)(touchCoords.pos2.x)].index = cInvalidCoordValue; return true; } // 若在同一行格子 if (touchCoords.pos1.y == touchCoords.pos2.y) { if ((touchCoords.pos1.y != 0) && (touchCoords.pos1.y != cGridRows - 1)) { if (touchCoords.pos1.x < touchCoords.pos2.x) { for (int i = (int)(touchCoords.pos1.x) + 1; i < ((int)(touchCoords.pos2.x)); i++) { if (genies[(int)(touchCoords.pos1.y), i].index != cInvalidCoordValue) { return false; } } } else { for (int i = (int)(touchCoords.pos2.x) + 1; i < ((int)(touchCoords.pos1.x)); i++) { if (genies[(int)(touchCoords.pos1.y), i].index != cInvalidCoordValue) { return false; } } } } genies[(int)(touchCoords.pos1.y), (int)(touchCoords.pos1.x)].index = cInvalidCoordValue; genies[(int)(touchCoords.pos2.y), (int)(touchCoords.pos2.x)].index = cInvalidCoordValue; return true; } return false; } // 1 个转角连通 --> 相当于两个格子划出一个矩形, 这两个格子是一对对角顶点, 另两个顶点如果可以同时和这两个格子直连, 那就说明可以连通 bool CheckLink1() { int l = cInvalidCoordValue, t = cInvalidCoordValue, r = cInvalidCoordValue, b = cInvalidCoordValue; if (touchCoords.pos1.y < touchCoords.pos2.y) { t = (int)(touchCoords.pos1.y); b = (int)(touchCoords.pos2.y); } else { t = (int)(touchCoords.pos2.y); b = (int)(touchCoords.pos1.y); } if (touchCoords.pos1.x < touchCoords.pos2.x) { l = (int)(touchCoords.pos1.x); r = (int)(touchCoords.pos2.x); } else { l = (int)(touchCoords.pos2.x); r = (int)(touchCoords.pos1.x); } if (genies[t, l].index == cInvalidCoordValue) { // 若选取的两个格子在 右上、左下 for (int i = t + 1; i < b; i++) { if (genies[i, l].index != cInvalidCoordValue) { return false; } } for (int j = l + 1; j < r; j++) { if (genies[t, j].index != cInvalidCoordValue) { return false; } } genies[t, r].index = cInvalidCoordValue; genies[b, l].index = cInvalidCoordValue; return true; } else if (genies[t, r].index == cInvalidCoordValue) { // 若选取的两个格子在 左上、右下 for (int i = t + 1; i < b; i++) { if (genies[i, r].index != cInvalidCoordValue) { return false; } } for (int j = l + 1; j < r; j++) { if (genies[t, j].index != cInvalidCoordValue) { return false; } } genies[t, l].index = cInvalidCoordValue; genies[b, r].index = cInvalidCoordValue; return true; } else if (genies[b, l].index == cInvalidCoordValue) { // 若选取的两个格子在 左上、右下 for (int i = t + 1; i < b; i++) { if (genies[i, r].index != cInvalidCoordValue) { return false; } } for (int j = l + 1; j < r; j++) { if (genies[b, j].index != cInvalidCoordValue) { return false; } } genies[t, l].index = cInvalidCoordValue; genies[b, r].index = cInvalidCoordValue; return true; } else if (genies[b, r].index == cInvalidCoordValue) { // 若选取的两个格子在 右上、左下 for (int i = t + 1; i < b; i++) { if (genies[i, l].index != cInvalidCoordValue) { return false; } } for (int j = l + 1; j < r; j++) { if (genies[b, j].index != cInvalidCoordValue) { return false; } } genies[t, r].index = cInvalidCoordValue; genies[b, l].index = cInvalidCoordValue; return true; } return false; }
游戏真机运行截图如下。
代码下载链接在这里。
FruitFrolic
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