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1、
2、题目大意：
给定一个二维的矩阵，其中的数分别代表一些高度，现在可以从某一点滑雪，只能从高处往低处滑，问一条路径最多可以经过多少个点
3、题目：
Problem C
Longest Run on a Snowboard
Input: standard input
Output: standard output
Time Limit: 5 seconds
Memory Limit: 32 MB
Michael likes snowboarding. That's not very surprising, since snowboarding is really great. The bad thing is that in order to gain speed, the area must slide downwards. Another disadvantage is that when you've reached the bottom of the hill you have to walk up again or wait for the ski-lift.
Michael would like to know how long the longest run in an area is. That area is given by a grid of numbers, defining the heights at those points. Look at this example:
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
One can slide down from one point to a connected other one if and only if the height decreases. One point is connected to another if it's at left, at right, above or below it. In the sample map, a possible slide would be 24-17-16-1 (start at 24, end at 1). Of course if you would go 25-24-23-...-3-2-1, it would be a much longer run. In fact, it's the longest possible.
Input
The first line contains the number of test cases N. Each test case starts with a line containing the name (it's a single string), the number of rows Rand the number of columns C. After that follow R lines with C numbers each, defining the heights. R and C won't be bigger than 100, N not bigger than 15 and the heights are always in the range from 0 to 100.
For each test case, print a line containing the name of the area, a colon, a space and the length of the longest run one can slide down in that area.
Sample Input
2
Feldberg 10 5
56 14 51 58 88
26 94 24 39 41
24 16 8 51 51
76 72 77 43 10
38 50 59 84 81
5 23 37 71 77
96 10 93 53 82
94 15 96 69 9
74 0 62 38 96
37 54 55 82 38
Spiral 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
Feldberg: 7
Spiral: 25
(Math Lovers’ Contest, Problem Setter: Stefan Pochmann)
4、代码：
#include#include #define N 105#include using namespace std;int a[N][N];int dp[N][N];int dir[4][2]={-1,0,1,0,0,-1,0,1};int m,n;int dfs(int x,int y){ int tmp=0; if(dp[x][y]) return dp[x][y]; for(int i=0;i<4;i++) { int tx=x+dir[i][0]; int ty=y+dir[i][1]; if(tx>=1 && tx<=m && ty>=1 && ty<=n && a[tx][ty]
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