只要注意一下细节就毫无难点了，简简单单状态压缩即可。
Code:
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 20;
const long long mod = 4921057 ;
int a[maxn], b[maxn], pos[maxn], sumv[maxn];
long long dp[1 << maxn][maxn];
inline int lowbit(int t) { return t & (-t); }
inline int dis(int x,int y,int S){
if(x < y) swap(x, y);
int ans = sumv[x - 1] - sumv[y];
for(int i = x; i > y; --i) if((S & pos[i]) != 0) --ans;
return ans;
}
int main()
{
int n;
scanf("%d",&n);
for(int i = 1;i <= n + 1; ++i) pos[i] = (1 << (i - 1));
for(int i = 1;i <= n; ++i) { scanf("%d",&a[i]); sumv[i] = sumv[i - 1] + 1; }
for(int i = 1;i <= n; ++i) scanf("%d",&b[i]);
for(int i = 1;i <= n; ++i) dp[pos[i]][i] = 1;
for(int i = 1;i < pos[n + 1]; ++i)
{
for(int j = 1;j <= n; ++j)
{
if((i & pos[j]) == 0) continue;
int mx = (i ^ pos[j]);
if(mx == 0) continue;
for(int k = 1;k <= n; ++k)
{
if(k == j || (mx & pos[k]) == 0 || a[j] < a[k]) continue;
if(dis(k, j,mx) > b[k]) continue;
dp[i][j] += dp[mx][k];
dp[i][j] %= mod;
}
}
}
long long fin = 0;
for(int i = 1;i <= n; ++i) fin += dp[pos[n + 1] - 1][i] , fin %= mod;
printf("%lld",fin);
return 0;
}