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 }