uva 10285 - Longest Run on a Snowboard（dp+记忆化搜索）

最新推荐文章于 2019-04-07 11:27:00 发布

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本文介绍了一种解决最长滑雪路径问题的算法实现。通过深度优先搜索(DFS)递归地寻找从任意点出发能到达的最大点数，最终找出全局最大值。文章提供了完整的C++代码示例，并详细解释了代码的各个部分。

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1、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1226

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

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<stdio.h>
#include<string.h>
#define N 105
#include<algorithm>
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]<a[x][y])
        {
            tmp=max(tmp,dfs(tx,ty));
        }
    }
    dp[x][y]=tmp+1;
    return dp[x][y];
}
int main()
{
    int t;
    char str[105];
    scanf("%d",&t);
    while(t--)
    {
        int ans=0;
        scanf("%s%d%d",str,&m,&n);
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            scanf("%d",&a[i][j]);
        }
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(dp[i][j])//如果计算过了
                ans=max(ans,dp[i][j]);
                else
                ans=max(ans,dfs(i,j));
            }
        }
        printf("%s: %d\n",str,ans);
    }
    return 0;
}