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
R
and 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)
题意：
Michael很喜欢滑雪。滑雪很好玩，但是有一点比较麻烦。就是为了要获得速度，滑雪一定要由高处往低处滑。等你到了山脚时就得走路上山或等待滑雪登山缆车了。
Michael想要知道在某一个滑雪场最长的滑雪路径有多长。滑雪场区域是以数字形成的方块来表示。数字的大小代表各个点的高度。看以下的例子：
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
我们可以从一点滑到相连的另一点，只要高度是由高到低。在这里我们说某一点与另一点相连指的是他们互为上、下、左、右四个方向相邻。在上面的地图中我们可以滑
24-17-16-1
（从24开始，1结束）。当然，假如你想要滑
25-24-23-22-….-3-2-1
也可以，这比上一条路径长多了。事实上，这也是最长的路径了。
思路：DAG上的动态规划，只是方向变成了上下左右四个方向，状态转移公式没有变化。
#include<iostream>
#include<cstring>
using namespace std;
int map[110][110],vis[110][110],arry[110][110];
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int n,m;
int dp(int x,int y)
{
if(vis[x][y]) return arry[x][y];
for(int i=0;i<4;i++)
{
int dx=x+dir[i][0],dy=y+dir[i][1];
if(dx>=0&&dx<n&&dy>=0&&dy<m&&map[dx][dy]<map[x][y])
{
arry[x][y]=max(dp(dx,dy)+1,arry[x][y]);
}
}
vis[x][y]=1;
return arry[x][y];
}
int main()
{
int num;
cin>>num;
while(num--)
{
string name;
cin>>name>>n>>m;
memset(map, 0, sizeof(map));
memset(vis, 0, sizeof(vis));
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
cin>>map[i][j];
arry[i][j]=0;
}
}
int maxn=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
int cnt=dp(i,j);
if(cnt>maxn) maxn=cnt;
}
}
cout<<name<<": "<<maxn+1<<endl;
}
return 0;
}