連連看遊戲的消除判定是3條直線之內能連上即可消除
那麼,要如何實現呢
首先,我們把棋盤的數據用一個數組來存下來
然後分析一下消除條件
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 11 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 11 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
實質上,我們需要考慮的地方只有一個十字形,即(x1~x2,y(0~max))和(x(0~max),y1~y2)這2快區域
因爲只有直線,而且不能被阻隔,所以我們要先找一個x1到x2這塊之內是空的地方,然後再用找到的y來和2個點之間的那2條線是否都是可以連上的,y1到y2同理可得
根據這個算法,代碼就出來了
bool CheckTheLink(POINT p1, POINT p2)
{
//UpdateGameData();
if (p1.x == p2.x)
{
if (p1.y - p2.y == 1 || p2.y - p1.y == 1)
{
return true;
}
}
if (p1.y == p2.y)
{
if (p1.x - p2.x == 1 || p2.x - p1.x == 1)
{
return true;
}
}
for (int i = 0; i < 11; i++)
{
bool key = true;
for (int j = (p1.y > p2.y ? p2.y : p1.y) + 1; j < (p1.y < p2.y ? p2.y : p1.y); j++)
{
if (gameData[i][j] != 0)
{
key = false;
break;
}
}
if (!key)
{
continue;
}
else
{
bool key1 = true, key2 = true;
for (int k = (p1.x < i ? p1.x : i); k <= (p1.x > i ? p1.x : i); k++)
{
if (gameData[k][p1.y] != 0)
{
if (k != p1.x)
{
key1 = false;
break;
}
}
}
if (!key1)
{
//break;
continue;
}
for (int k = (p2.x < i ? p2.x : i); k <= (p2.x > i ? p2.x : i); k++)
{
if (gameData[k][p2.y] != 0)
{
if (k != p2.x)
{
key2 = false;
break;
}
}
}
if (key2)
{
return true;
}
}
}
for (int j = 0; j < 19; j++)
{
bool key = true;
for (int i = (p1.x > p2.x ? p2.x : p1.x) + 1; i < (p1.x < p2.x ? p2.x : p1.x); i++)
{
if (gameData[i][j] != 0)
{
key = false;
break;
}
}
if (!key)
{
continue;
}
else
{
bool key1 = true, key2 = true;
for (int k = (j > p1.y ? p1.y : j); k <= (j < p1.y ? p1.y : j); k++)
{
if (gameData[p1.x][k] != 0)
{
if (k != p1.y)
{
key1 = false;
break;
}
}
}
if (!key1)
{
//break;
continue;
}
for (int k = (j > p2.y ? p2.y : j); k <= (j < p2.y ? p2.y : j); k++)
{
if (gameData[p2.x][k] != 0)
{
if (k != p2.y)
{
key2 = false;
break;
}
}
}
if (key2)
{
return true;
}
}
}
return false;
}
棋盤是19*11的,可以根據實際需求更改