滑雪
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 88484   Accepted: 33177

Description

Michael 喜欢滑雪百这并不奇怪， 因为滑雪的确很刺激。可是为了获得速度，滑的区域必须向下倾斜，而且当你滑到坡底，你不得不再次走上坡或者等待升降机来载你。Michael想知道载一个 区域中最长底滑坡。区域由一个二维数组给出。数组的每个数字代表点的高度。下面是一个例子 
 1  2  3  4 5

16 17 18 19 6

15 24 25 20 7

14 23 22 21 8

13 12 11 10 9

一个人可以从某个点滑向上下左右相邻四个点之一，当且仅当高度减小。在上面的例子中，一条可滑行的滑坡为24-17-16-1。当然25-24-23-...-3-2-1更长。事实上，这是最长的一条。

Input

输入的第一行表示区域的行数R和列数C(1 <= R,C <= 100)。下面是R行，每行有C个整数，代表高度h，0<=h<=10000。

Output

输出最长区域的长度。

Sample Input

5 5
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9

Sample Output

25

分析：看到这个题感觉好简单，直接递归就写好了，提交居然TLE了。那么就需要一些小技巧了，先干掉动态分配二维数组，然后定义一个二维数组gawDis[]用来记录对应节点的最长路径，后面再用到这个节点的时候就不需要再计算了。
综上：递归+记忆化dfs搜索
  1 #include <stdio.h>
  2 #include <stdlib.h>
  3 
  4 #define MAXNUM 101
  5 
  6 int gawMap[MAXNUM][MAXNUM];
  7 int gawRow;
  8 int gawColumn;
  9 int gawDis[MAXNUM][MAXNUM];
 10 
 11 int max(int a, int b)
 12 {
 13     if(a>b)
 14         return a;
 15     else
 16         return b;
 17 }
 18 
 19 int Ski(int wlasthigh, int *awlastposition)
 20 {
 21     int awposition[2] = {0};
 22     int wleft = 1;
 23     int wright = 1;
 24     int wup = 1;
 25     int wdown = 1;
 26 
 27     if(gawDis[awlastposition[0]][awlastposition[1]] != 0)
 28     {
 29         return gawDis[awlastposition[0]][awlastposition[1]];
 30     }
 31     if(awlastposition[1] > 0)
 32     {
 33         awposition[1] = awlastposition[1]-1;
 34         awposition[0] = awlastposition[0];
 35         if(wlasthigh > gawMap[awposition[0]][awposition[1]])
 36         {
 37             wleft = Ski(gawMap[awposition[0]][awposition[1]], awposition) + 1;
 38         }
 39     }
 40     if(awlastposition[1] < gawColumn-1)
 41     {
 42         awposition[1] = awlastposition[1]+1;
 43         awposition[0] = awlastposition[0];
 44         if(wlasthigh > gawMap[awposition[0]][awposition[1]])
 45         {
 46             wright = Ski(gawMap[awposition[0]][awposition[1]], awposition) + 1;
 47         }
 48     }
 49     if(awlastposition[0] > 0)
 50     {
 51         awposition[0] = awlastposition[0]-1;
 52         awposition[1] = awlastposition[1];
 53         if(wlasthigh > gawMap[awposition[0]][awposition[1]])
 54         {
 55             wup = Ski(gawMap[awposition[0]][awposition[1]], awposition) + 1;
 56         }
 57     }
 58     if(awlastposition[0] < gawRow-1)
 59     {
 60         awposition[0] = awlastposition[0]+1;
 61         awposition[1] = awlastposition[1];
 62         if(wlasthigh > gawMap[awposition[0]][awposition[1]])
 63         {
 64             wdown = Ski(gawMap[awposition[0]][awposition[1]], awposition) + 1;
 65         }
 66     }
 67     wleft = max(wleft, wright);
 68     wup = max(wup, wdown);
 69     return gawDis[awlastposition[0]][awlastposition[1]] = max(wleft, wup);
 70 }
 71 
 72 int main(void)
 73 {
 74     int i = 0;
 75     int j = 0;
 76     scanf("%d %d", &gawRow, &gawColumn);
 77 
 78     for(i=0; i<gawRow; i++)
 79     {
 80         for(j=0; j<gawColumn; j++)
 81         {
 82             scanf("%d", &gawMap[i][j]);
 83         }
 84     }
 85 
 86     int awposition[2] = {0};
 87     for(i=0; i<gawRow; i++)
 88     {
 89         for(j=0; j<gawColumn; j++)
 90         {
 91             awposition[1] = j;
 92             awposition[0] = i;
 93             Ski(gawMap[awposition[0]][awposition[1]], awposition);
 94         }
 95     }
 96 
 97     int wLen = 0;
 98     for(i=0; i<gawRow; i++)
 99     {
100         for(j=0; j<gawColumn; j++)
101         {
102             if(wLen < gawDis[i][j]) wLen = gawDis[i][j];
103         }
104     }
105     printf("%d\n", wLen);
106 
107     return 0;
108 }