極大化思想的應用！
用l，r分別記錄了之前的狀態！
下一次查找的時候只需要o(1)的時間，最壞情況下是o(n),不過由于壞點很稀疏，最壞情況很少發生！
所以程序非常快！

   Test 1: TEST OK [0.000 secs, 11544 KB]
   Test 2: TEST OK [0.011 secs, 11544 KB]
   Test 3: TEST OK [0.011 secs, 11544 KB]
   Test 4: TEST OK [0.011 secs, 11544 KB]
   Test 5: TEST OK [0.011 secs, 11544 KB]
   Test 6: TEST OK [0.022 secs, 11540 KB]
   Test 7: TEST OK [0.054 secs, 11544 KB]
   Test 8: TEST OK [0.194 secs, 11544 KB]
   Test 9: TEST OK [0.302 secs, 11540 KB]
   Test 10: TEST OK [0.281 secs, 11544 KB]

 1 動態規劃
 2 小炫耀一下
 3 程序運行的非常快么！
 4 /*
 5 ID: hongfei5
 6 PROG: rectbarn
 7 LANG: C++
 8 */
 9 #include<iostream>
10 using namespace std;
11 bool map[3001][3001];
12 int h[3001],l[3001],r[3001];
13 int n,m,p;
14 int main()
15 {
16       freopen("rectbarn.in","r",stdin);
17       freopen("rectbarn.out","w",stdout);
18       scanf("%d%d%d",&n,&m,&p);
19       int a,b;
20       for(int i=0;i<p;++i)
21       {
22             scanf("%d%d",&a,&b);
23             map[a][b]=1;//1表示是壞點
24       }
25       h[0]=0;
26       int Max=0; 
27       //[i,j]為(i,h[i,j])這條線段向左邊擴展的最長距離，r[i,j]為(i,h[i,j])向右邊擴展的最長距離
28       for(int i=1;i<=n;++i)
29       {
30             l[i]=0;
31             r[i]=m+1;
32             for(int j=1;j<=m;++j)
33                   if(map[i][j])
34                   {
35                         r[i]=j;
36                         break;
37                   }
38       }
39       for(int j=1;j<=m;++j)
40       {
41             int len_l=INT_MAX;
42             int len_r=INT_MAX;
43             for(int i=1;i<=n;++i)
44             { 
45                   if(map[i][j])
46                   {
47                         l[i]=j;
48                         h[i]=0;
49                         r[i]=m+1;
50                         len_l=len_r=INT_MAX;
51                         for(int k=j+1;k<=m;++k)
52                               if(map[i][k])
53                               {
54                                     r[i]=k;
55                                     break;
56                               }
57                   }
58                   else 
59                   {
60                         h[i]=h[i-1]+1;
61                         len_l=min(len_l,j-l[i]);
62                         len_r=min(len_r,r[i]-j);
63                         int s=(len_l+len_r-1)*h[i];
64                         if(s>Max)Max=s;
65                   }
66             }
67       }
68       printf("%d\n",Max); 
69       return 0;
70 }