[UVa 10285]Longest Run on a Snowboard

Problem Description

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.

Output

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

Problem Report

一道简单的记忆化搜索题。因为题目要求严格递减，所以不用担心在一次搜索过程中走到已经走过的点上，因此可以在回溯的时候保存以当前点出发能走的最长路。

My Source Code

//  Created by Chlerry in 2015.
//  Copyright (c) 2015 Chlerry. All rights reserved.
//

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cstring>
#include <climits>
#include <string>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define Size 120
#define ll long long
#define mk make_pair
#define pb push_back
#define mem(array) memset(array,0,sizeof(array))
typedef pair<int,int> P;

int n,m,a[Size][Size],d[Size][Size],maxn;
const int addx[]={0,0,-1,1};
const int addy[]={-1,1,0,0};
#define tx x+addx[i]
#define ty y+addy[i]

int DFS(int x,int y)
{
    if(d[x][y])
        return d[x][y];
    for(int i=0;i<4;i++) if(tx>=0 && tx<n && ty>=0 && ty<m && a[tx][ty]<a[x][y])
        d[x][y]=max(d[x][y],DFS(tx,ty));
    return ++d[x][y];
}
int main()
{
    freopen("in.txt","r",stdin);
    int T;cin>>T;
while(T--)
{
    string name;cin>>name>>n>>m;maxn=INT_MIN;
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
            cin>>a[i][j],d[i][j]=0;
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
            maxn=max(maxn,DFS(i,j));//,cout<<d[i][j]<<" ";
    cout<<name<<": "<<maxn<<endl;
}

    return 0;
}