玩家鼠標先後點擊兩塊棋子,試圖將他們消去,然後遊戲的後臺判斷這兩個方格能不能消去。現在你的任務就是寫這個後臺程序。
注意:詢問之間無先後關係,都是針對當前狀態的!
- #include
- #include
- #include
- using namespace std;
- #define maxn 1005
- int map[maxn][maxn];
- int d[maxn][maxn];
- int m, n, x2, y2;
- int dir[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};
- struct node
- {
- int k, ways, x, y;
- friend bool operator < (node a, node b)
- {
- return a.k > b.k;
- }
- };
-
- int bfs(int x1, int y1);
- void clean();
-
- int main()
- {
- while(scanf("%d%d", &m, &n), m || n)
- {
- for(int i=1; i<=m; i++)
- for(int j=1; j<=n; j++)
- scanf("%d", &map[i][j]);
-
- int T, x1, y1;
-
- scanf("%d", &T);
-
- while(T--)
- {
- scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
-
- if(!map[x1][y1] || map[x1][y1]!=map[x2][y2] || x1==x2&&y1==y2)
- {
- printf("NO\n");
- continue;
- }
- clean();
- int res = bfs(x1, y1);
-
- if(res)
- printf("YES\n");
- else
- printf("NO\n");
- }
- }
-
- return 0;
- }
- int bfs(int x1, int y1)
- {
- priority_queue que;
- node q, s;
- d[x1][y1] = q.k = -1, q.ways = -1, q.x = x1, q.y = y1;
- que.push(q);
-
- while(que.size())
- {
- q = que.top(), que.pop();
-
- if(q.k == 3)break;
- if(q.x == x2 && q.y == y2)return 1;
-
- for(int i=0; i<4; i++)
- {
- s.x = q.x + dir[i][0], s.y = q.y + dir[i][1];
-
- if(s.x>0&&s.x<=m && s.y>0&&s.y<=n && !map[s.x][s.y] || s.x==x2 && s.y==y2)
- {
- if(q.ways != i)s.k = q.k + 1;
- else s.k = q.k;
- s.ways = i;
-
- if(s.k <= d[s.x][s.y])
- {
- d[s.x][s.y] = s.k;
- que.push(s);
- }
- }
- }
- }
-
- return 0;
- }
- void clean()
- {
- for(int i=0; i<=m; i++)
- for(int j=0; j<=n; j++)
- d[i][j] = 10;
- }