ZOJ 2859 Matrix Searching

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1859

题目：给出一个矩阵，求出指定子矩阵中的最小元素。

我用的不是正规解法，纯属水过去的……这个是留着自己看的，大家就不要学了。

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 const int MAXN = 300 + 10;
 5 
 6 int map[MAXN][MAXN];
 7 int Mat[MAXN][MAXN];
 8 
 9 int n;
10 int INF = 2147483645;
11 
12 int min( int a, int b )
13 {
14     return a < b ? a : b;
15 }
16 
17 int Findmin( int x1, int y1, int x2, int y2 )
18 {
19     if ( x1 == 1 && y1 == 1 ) return Mat[x2][y2];
20     if ( Mat[x2][y2] != Mat[x1][y2] && Mat[x2][y2] != Mat[x2][y1] ) return Mat[x2][y2];
21 
22     int mmin = INF;
23     for ( int i = x1; i <= x2; i++ )
24        for ( int j = y1; j <= y2; j++ )
25           if ( map[i][j] < mmin ) mmin = map[i][j];
26 
27     return mmin;
28 }
29 
30 int main()
31 {
32     int T;
33     scanf( "%d", &T );
34     while ( T-- )
35     {
36         scanf( "%d", &n );
37 
38         for ( int i = 1; i <= n; i++ )
39         {
40             for ( int j = 1; j <= n; j++ )
41             {
42                 scanf( "%d", &map[i][j] );
43                 if ( i == 1 && j == 1 )
44                     Mat[i][j] = map[i][j];
45                 else if ( i != 1 && j == 1 )
46                     Mat[i][j] = min( map[i][j], Mat[i - 1][j] );
47                 else if ( i == 1 && j != 1 )
48                     Mat[i][j] = min( map[i][j], Mat[i][j - 1] );
49                 else Mat[i][j] = min( map[i][j], min(Mat[i - 1][j], Mat[i][j - 1]) );
50             }
51         }
52 
53         int m;
54         scanf( "%d", &m );
55         while ( m-- )
56         {
57             int a, b, c, d;
58             scanf("%d%d%d%d", &a, &b, &c, &d);
59             printf("%d\n", Findmin( a, b, c, d ) );
60         }
61     }
62     return 0;
63 }