滑雪

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://www.tyvj.cn/p/1004

Description

Input

Output

Sample Input

Sample Output

HINT

题意

题解：

啊，记忆化搜索，BFS和DFS随便乱搞都行，注意得把所有点都扔进队列里面

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 2001

#define mod 10007

#define eps 1e-9

//const int inf=0x7fffffff;   //无限大

const int inf=0x3f3f3f3f;

/*

inline ll read()

{

    int x=0,f=1;char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

int buf[10];

inline void write(int i) {

  int p = 0;if(i == 0) p++;

  else while(i) {buf[p++] = i % 10;i /= 10;}

  for(int j = p-1; j >=0; j--) putchar('0' + buf[j]);

  printf("\n");

}

*/

//**************************************************************************************

inline ll read()

{

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int g[maxn][maxn];

int dp[maxn][maxn];

int dx[]={,-,,};

int dy[]={,,,-};

int main()

{

    int mx=;

    int my=;

    int ma=;

    int n=read(),m=read();

    queue<int> qx;

    queue<int> qy;

    for(int i=;i<=n;i++)

    {

        for(int j=;j<=m;j++)

        {

            g[i][j]=read();

            dp[i][j]=;

            qx.push(i);

            qy.push(j);

        }

    }

    while(!qx.empty())

    {

        int nowx=qx.front();

        int nowy=qy.front();

        qx.pop();

        qy.pop();

        for(int i=;i<;i++)

        {

            int nextx=nowx+dx[i];

            int nexty=nowy+dy[i];

            if(nextx<||nextx>n||nexty<||nexty>m)

                continue;

            if(g[nextx][nexty]<g[nowx][nowy])

            {

                if(dp[nowx][nowy]>=dp[nextx][nexty])

                {

                    dp[nextx][nexty]=dp[nowx][nowy]+;

                    qx.push(nextx);

                    qy.push(nexty);

                }

            }

        }

    }

    int ans=;

    for(int i=;i<=n;i++)

    {

        for(int j=;j<=m;j++)

        {

            ans=max(dp[i][j],ans);

        }

    }

    cout<<ans<<endl;

}