题目：vj地址
思路：dp[i][j][k]代表 以i，j结尾 有k长度的路径数量，k最大等于4，如果k超过4，也是等于4.
那么转移：dp[i][j][k]+={dp[x][y][k-1]}(x,y满足a[x][y]+1=a[i][j]);
如果k=4,还有dp[i][j][k]+={dp[x][y][k]}(x,y满足a[x][y]+1=a[i][j]);

细节见代码。

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define FILL(a,b) (memset(a,b,sizeof(a)))
#define re register
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(a) ((a)&-(a))
#define ios std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);
#define fi first
#define rep(i,n) for(int i=0;(i)<(n);i++)
#define rep1(i,n) for(int i=1;(i)<=(n);i++)
#define se secondusing namespace std;
typedef long long  ll;
typedef unsigned long long  ull;
typedef pair<int,int > pii;
int dx[4]= {-1,1,0,0},dy[4]= {0,0,1,-1};
const ll mod=1e9+7;
const ll N =2e6+10;
const double eps = 1e-4;
const double pi=acos(-1);
ll gcd(int a,int b){return !b?a:gcd(b,a%b);}
ll dp[1100][1100][5];
ll a[1100][1100];
ll n,m;
bool check1(ll x,ll y)//检查是否能转移
{for(ll i=0;i<4;i++){ll x1=x+dx[i],y1=y+dy[i];if(x1<=0||x1>n||y1<=0||y1>m) continue;if(a[x][y]-1==a[x1][y1]) return 0;}return 1;
}
bool check2(ll x,ll y)
{for(ll i=0;i<4;i++){ll x1=x+dx[i],y1=y+dy[i];if(x1<=0||x1>n||y1<=0||y1>m) continue;if(a[x][y]+1==a[x1][y1]) return 0;}return 1;
}
ll dfs(ll i,ll j,ll k)//记忆化搜索
{if(k==1){if(check1(i,j)) return 1;else return 0;}if(dp[i][j][k]!=-1) return dp[i][j][k];ll &ans=dp[i][j][k];ans=0;for(ll w=0;w<4;w++){ll x1=i+dx[w],y1=j+dy[w];if(x1<=0||x1>n||y1<=0||y1>m) continue;if(a[x1][y1]+1==a[i][j]){ans+=dfs(x1,y1,k-1);if(k==4) ans=(ans+dfs(x1,y1,4))%mod;//k==4时的特判转移条件}}return ans;
}
int main()
{#ifdef LOCALfreopen("in.txt", "r", stdin);#endifmemset(dp,-1,sizeof dp);scanf("%lld%lld",&n,&m);for(int i=1;i<=n;i++)for(int j=1;j<=m;j++) scanf("%lld",&a[i][j]);ll ans=0;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(check2(i,j)) ans=(ans+dfs(i,j,4))%mod;//满足转移条件时才能加}}cout<<ans<<endl;return 0;
}