考虑dp。
对于ai，和他能匹配的bj只有9个，所以我们考虑从这9个状态转移。
对于ai 能匹配的一个bj，当前最大的匹配数一定是[1, j - 1]中的最大匹配数 + 1。然后用树状数组维护前缀匹配数最大值就行了。
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxn = 1e5 + 5; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 27 if(last == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) x = -x, putchar('-'); 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 int n, a[maxn], pos[maxn]; 38 int Max[maxn]; 39 40 int c[maxn]; 41 int lowbit(int x) 42 { 43 return x & -x; 44 } 45 void add(int pos, int x) 46 { 47 for(; pos <= n && c[pos] < x; pos += lowbit(pos)) c[pos] = x; 48 } 49 int query(int pos) 50 { 51 int ret = 0; 52 for(; pos; pos -= lowbit(pos)) if(c[pos] > ret) ret = c[pos]; 53 return ret; 54 } 55 56 int main() 57 { 58 n = read(); 59 for(int i = 1; i <= n; ++i) a[i] = read(); 60 for(int i = 1; i <= n; ++i) {int x = read(); pos[x] = i;} 61 for(int i = 1; i <= n; ++i) 62 { 63 for(int j = max(1, a[i] - 4); j <= min(n, a[i] + 4); ++j) 64 Max[j] = query(pos[j] - 1); 65 for(int j = max(1, a[i] - 4); j <= min(n, a[i] + 4); ++j) 66 add(pos[j], Max[j] + 1); 67 } 68 write(query(n)), enter; 69 return 0; 70 }