BFS 2015百度之星初赛2 HDOJ 5254 棋盘占领

 

题目传送门

 1 /*  2  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  3  注意：结果还要加上原来占领的  4 */  5 #include <cstdio>  6 #include <algorithm>  7 #include <cstring>  8 #include <cmath>  9 #include <vector>  10 #include <queue>  11 #include <set>  12 using namespace std;  13  14 const int MAXN = 5e2 + 10;  15 const int INF = 0x3f3f3f3f;  16 int a[MAXN][MAXN];  17 int dx[4] = {-1, -1, 1, 1};  18 int dy[4] = {-1, 1, -1, 1};  19 int n, m;  20 queue<pair<int, int> > Q;  21  22 int change(int x, int y)  23 {  24  25 int cnt = 0;  26 for (int i=0; i<4; ++i)  27  {  28 int tx = x + dx[i]; int ty = y + dy[i];  29 if (tx < 1 || tx > n) continue;  30 if (ty < 1 || ty > m) continue;  31 if (a[tx][ty] == 1)  32  {  33 if (i == 0 || i == 2)  34  {  35 if (a[tx][ty+1] == 0) {a[tx][ty+1] = 1; ++cnt; Q.push (make_pair (tx, ty+1));}  36 if (a[x][y-1] == 0) {a[x][y-1] = 1; ++cnt; Q.push (make_pair (x, y-1));}  37  }  38 else if (i == 1 || i == 3)  39  {  40 if (a[tx][ty-1] == 0) {a[tx][ty-1] = 1; ++cnt; Q.push (make_pair (tx, ty-1));}  41 if (a[x][y+1] == 0) {a[x][y+1] = 1; ++cnt; Q.push (make_pair (x, y+1));}  42  }  43  }  44  }  45  46 return cnt;  47 }  48  49 int BFS(void)  50 {  51 int ans = 0;  52 while (!Q.empty ())  53  {  54 int x = Q.front ().first; int y = Q.front ().second; Q.pop ();  55 ans += change (x, y);  56  }  57  58 return ans;  59 }  60  61 int main(void) //2015百度之星初赛2 HDOJ 5254 棋盘占领  62 {  63 int t, cas = 0; scanf ("%d", &t);  64 while (t--)  65  {  66 while (!Q.empty ()) Q.pop ();  67 scanf ("%d%d", &n, &m);  68 memset (a, 0, sizeof (a));  69  70 int g; scanf ("%d", &g);  71 int cnt = 0;  72 while (g--)  73  {  74 int x, y; scanf ("%d%d", &x, &y);  75 if (a[x][y] == 0)  76  {  77 cnt++; a[x][y] = 1; Q.push (make_pair (x, y));  78  }  79  }  80  81 printf ("Case #%d:\n", ++cas);  82 printf ("%d\n", BFS () + cnt);  83  }  84  85 return 0;  86 }  87  88  89  90  91 /*  92 4  93 2 2  94 2  95 1 1  96 2 2  97 3 3  98 3  99 1 1 100 2 3 101 3 2 102 2 4 103 5 104 1 1 105 1 1 106 1 2 107 1 3 108 1 4 109 2 4 110 2 111 1 1 112 2 4 113 */