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:
", ++cas);
82 printf ("%d
", 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 */