uva 10285 Longest Run on a Snowboard

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

这道题目还是很好理解的，使用DFS可以强行解之，但是需要注意的地方就是用一个数组存储已经DFS过的节点的最大长度，所以思路也就很容易了，注意一点格式，就可以迅速解决。

#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#define max(a,b) (((a)>(b))?(a):(b))

using namespace std;

const int maxn = 105;
int n,m;
int G[maxn][maxn];
int dp[maxn][maxn];
int f[4][2] = {
    -1,0,1,0,0,1,0,-1
};

int dfs (int i , int j , int l) {
    int res = l;
    for (int s = 0 ; s < 4 ; s++) {
        int tx = i+f[s][0];
        int ty = j+f[s][1];
        if (tx >= 1 && tx <= n && ty >= 1 && ty <= m && G[i][j] > G[tx][ty]) {
            if (dp[tx][ty]) {
                res += dp[tx][ty];
                continue;
            }else 
                res = max(res,dfs(tx,ty,l+1));
        }
    }  
    return res;
}

int solve () {
    int res = 0;
    for (int i = 1 ; i <= n ; i++) {
        for (int j = 1 ; j <= m ; j++) {
            if (dp[i][j])
                res = max(res,dp[i][j]);
            else
                res = max(res,dfs(i,j,1));
        }
    }
    return res;
}

int main () {
    int cnt;
    cin >> cnt;
    for (int k = 0 ; k < cnt ; k++) {
        string tmp;
        cin >> tmp >> n >> m;
        memset (dp,0,sizeof(dp));
        memset (G,0,sizeof(G));
        for (int i = 1 ; i <= n ; i++) 
            for (int j = 1 ; j <= m ; j++) 
                scanf("%d" , &G[i][j]);
        cout << tmp << ": " << solve() << endl;
    }
}

经过别人解题报告的启发，我觉得下面这个修改过的代码更加优秀。虽然只是修改了一点，但是在思想上和运行速度上确实更加的优秀。下面是更新版代码。

#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#define max(a,b) (((a)>(b))?(a):(b))

using namespace std;

const int maxn = 105;
int n,m;
int G[maxn][maxn];
int dp[maxn][maxn];
int f[4][2] = {
    -1,0,1,0,0,1,0,-1
};

int dfs (int i , int j) {
    if (dp[i][j])
        return dp[i][j];
    int res = 1;
    for (int s = 0 ; s < 4 ; s++) {
        int tx = i+f[s][0];
        int ty = j+f[s][1];
        if (tx >= 1 && tx <= n && ty >= 1 && ty <= m && G[i][j] > G[tx][ty]) { 
            res = max(res,dfs(tx,ty)+1);
        }
    }  
    return dp[i][j] = res;
}

int solve () {
    int res = 0;
    for (int i = 1 ; i <= n ; i++) {
        for (int j = 1 ; j <= m ; j++) {
            if (dp[i][j])
                res = max(res,dp[i][j]);
            else
                res = max(res,dfs(i,j));
        }
    }
    return res;
}

int main () {
    int cnt;
    cin >> cnt;
    for (int k = 0 ; k < cnt ; k++) {
        string tmp;
        cin >> tmp >> n >> m;
        memset (dp,0,sizeof(dp));
        memset (G,0,sizeof(G));
        for (int i = 1 ; i <= n ; i++) 
            for (int j = 1 ; j <= m ; j++) 
                scanf("%d" , &G[i][j]);
        cout << tmp <<