滑雪问题

给定一个矩阵,要求找出数值递减的最长序列,比如矩阵:
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   最长的数值递减序列是25,24,23,22........1, 长度 25

问题满足后无效性，以某个数字结尾的最长序列只和以四周数字结尾的最长序列有关，和数组最长子序列类似，只不过最长子序列可以不连续，那么最长子序列的之前的子问题应该就是所有已经有解的数字。

// 1002.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <stdio.h>
#include <memory.h>
#include <queue>

using namespace std;

static const int MAX_ROW=100;
static const int MAX_COL=100;

int actualRow = MAX_ROW;
int actualCol = MAX_COL;

int gMatrix[MAX_ROW][MAX_COL] = {0};
int gPathLen[MAX_ROW*MAX_COL] = {0};

typedef struct _node
{
    int row;
    int col;
    int value;
    _node(int i,int j, int v)
    {
        row = i;
        col = j;
        value = v;
    };
    _node()
    {};
}Node;

struct _greater        
{    
    bool operator()(const Node& _Left, const Node& _Right) const
    {    
        return (_Left.value > _Right.value);
    }
};

int getMaxAdjLen(const Node& current)
{
    int result = 0;

    if (current.row > 0)
    {
        result = gPathLen[(current.row-1)*actualCol + current.col] > result ? 
                 gPathLen[(current.row-1)*actualCol + current.col] : result;
    }

    if (current.row < actualRow)
    {
        result = gPathLen[(current.row+1)*actualCol + current.col] > result ? 
                 gPathLen[(current.row+1)*actualCol + current.col] : result;
    }
    if (current.col > 0)
    {
        result = gPathLen[current.row*actualCol + current.col -1 ] > result ? 
                gPathLen[current.row*actualCol + current.col -1] : result;
    }
    if (current.col < actualCol)
    {
        result = gPathLen[current.row*actualCol + current.col +1 ] > result ? 
                gPathLen[current.row*actualCol + current.col +1] : result;
    }

    return result;
};
int main()
{
    int ar;
    int ac;    
    int maxLen ;
    Node tmp;
    priority_queue<Node,vector<Node>,_greater> Queue;
    while(scanf("%d%d",&ar,&ac) == 2)
    {
        actualCol = ac;
        actualRow = ar;
        maxLen = 0;
        memset(gPathLen,0,sizeof(int)*MAX_ROW*MAX_COL);
        memset(gMatrix,0,sizeof(int)*MAX_ROW*MAX_COL);
        for(int i=0;i<ar;i++)
            for(int j=0;j<ac;j++)
            {
                scanf("%d",&gMatrix[i][j]);
                Queue.push(Node(i,j,gMatrix[i][j]));
            }
while(!Queue.empty())
       {
           tmp = Queue.top();
           Queue.pop();
               gPathLen[tmp.row*ac+tmp.col] = getMaxAdjLen(tmp) + 1;
           maxLen = gPathLen[tmp.row*ac+tmp.col] > maxLen ? gPathLen[tmp.row*ac+tmp.col] : maxLen;
       }

           printf("%d",maxLen);
           while(!Queue.empty())
           Queue.pop();
    }
    return 0;
};