北大poj-1088

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

 #include <stdio.h>
 #include <stdlib.h>

 #define MAXNUM 101

 int gawMap[MAXNUM][MAXNUM];
 int gawRow;
 int gawColumn;
 int gawDis[MAXNUM][MAXNUM];

 int max(int a, int b)
 {
     if(a>b)
         return a;
     else
         return b;
 }

 int Ski(int wlasthigh, int *awlastposition)
 {
     int awposition[] = {};
     int wleft = ;
     int wright = ;
     int wup = ;
     int wdown = ;

     if(gawDis[awlastposition[]][awlastposition[]] != )
     {
         return gawDis[awlastposition[]][awlastposition[]];
     }
     if(awlastposition[] > )
     {
         awposition[] = awlastposition[]-;
         awposition[] = awlastposition[];
         if(wlasthigh > gawMap[awposition[]][awposition[]])
         {
             wleft = Ski(gawMap[awposition[]][awposition[]], awposition) + ;
         }
     }
     if(awlastposition[] < gawColumn-)
     {
         awposition[] = awlastposition[]+;
         awposition[] = awlastposition[];
         if(wlasthigh > gawMap[awposition[]][awposition[]])
         {
             wright = Ski(gawMap[awposition[]][awposition[]], awposition) + ;
         }
     }
     if(awlastposition[] > )
     {
         awposition[] = awlastposition[]-;
         awposition[] = awlastposition[];
         if(wlasthigh > gawMap[awposition[]][awposition[]])
         {
             wup = Ski(gawMap[awposition[]][awposition[]], awposition) + ;
         }
     }
     if(awlastposition[] < gawRow-)
     {
         awposition[] = awlastposition[]+;
         awposition[] = awlastposition[];
         if(wlasthigh > gawMap[awposition[]][awposition[]])
         {
             wdown = Ski(gawMap[awposition[]][awposition[]], awposition) + ;
         }
     }
     wleft = max(wleft, wright);
     wup = max(wup, wdown);
     return gawDis[awlastposition[]][awlastposition[]] = max(wleft, wup);
 }

 int main(void)
 {
     int i = ;
     int j = ;
     scanf("%d %d", &gawRow, &gawColumn);

     for(i=; i<gawRow; i++)
     {
         for(j=; j<gawColumn; j++)
         {
             scanf("%d", &gawMap[i][j]);
         }
     }

     int awposition[] = {};
     for(i=; i<gawRow; i++)
     {
         for(j=; j<gawColumn; j++)
         {
             awposition[] = j;
             awposition[] = i;
             Ski(gawMap[awposition[]][awposition[]], awposition);
         }
     }

     int wLen = ;
     for(i=; i<gawRow; i++)
     {
         for(j=; j<gawColumn; j++)
         {
             if(wLen < gawDis[i][j]) wLen = gawDis[i][j];
         }
     }
     printf("%d\n", wLen);

     return ;
 }