BFS 2015百度之星初赛2 HDOJ 5254 棋盘占领

题目传送门

 /*
     BFS：先把1的入队，每个1和它相邻的组合后看看能不能使0变1，若有则添加入队，change函数返回改变了多少个0
     注意：结果还要加上原来占领的
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 #include <vector>
 #include <queue>
 #include <set>
 using namespace std;

 const int MAXN = 5e2 + ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN][MAXN];
 int dx[] = {-, -, , };
 int dy[] = {-, , -, };
 int n, m;
 queue<pair<int, int> > Q;

 int change(int x, int y)
 {

     int cnt = ;
     for (int i=; i<; ++i)
     {
         int tx = x + dx[i];    int ty = y + dy[i];
         if (tx <  || tx > n)    continue;
         if (ty <  || ty > m)    continue;
         if (a[tx][ty] == )
         {
              if (i ==  || i == )
              {
                 if (a[tx][ty+] == )    {a[tx][ty+] = ;    ++cnt;    Q.push (make_pair (tx, ty+));}
                 if (a[x][y-] == )    {a[x][y-] = ;    ++cnt;    Q.push (make_pair (x, y-));}
              }
              else if (i ==  || i == )
              {
                 if (a[tx][ty-] == )    {a[tx][ty-] = ;    ++cnt;    Q.push (make_pair (tx, ty-));}
                 if (a[x][y+] == )    {a[x][y+] = ;    ++cnt;    Q.push (make_pair (x, y+));}
              }
         }
     }

     return cnt;
 }

 int BFS(void)
 {
     int ans = ;
     while (!Q.empty ())
     {
         int x = Q.front ().first;    int y = Q.front ().second;    Q.pop ();
         ans += change (x, y);
     }

     return ans;
 }

 int main(void)        //2015百度之星初赛2 HDOJ 5254 棋盘占领
 {
     int t, cas = ;    scanf ("%d", &t);
     while (t--)
     {
         while (!Q.empty ())    Q.pop ();
         scanf ("%d%d", &n, &m);
         memset (a, , sizeof (a));

         int g;    scanf ("%d", &g);
         int cnt = ;
         while (g--)
         {
             int x, y;    scanf ("%d%d", &x, &y);
             if (a[x][y] == )
             {
                 cnt++;    a[x][y] = ;    Q.push (make_pair (x, y));
             }
         }

         printf ("Case #%d:\n", ++cas);
         printf ("%d\n", BFS () + cnt);
     }

     return ;
 }

 /*
 4
 2 2
 2
 1 1
 2 2
 3 3
 3
 1 1
 2 3
 3 2
 2 4
 5
 1 1
 1 1
 1 2
 1 3
 1 4
 2 4
 2
 1 1
 2 4
 */