[USACO17FEB]Why Did the Cow Cross the Road II P

嘟嘟嘟

考虑dp。

对于a_i，和他能匹配的b_j只有9个，所以我们考虑从这9个状态转移。

对于a_i 能匹配的一个b_j，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。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = 1e5 + ;
 inline ll read()
 {
   ll ans = ;
   char ch = getchar(), last = ' ';
   while(!isdigit(ch)) {last = ch; ch = getchar();}
   while(isdigit(ch)) {ans = ans *  + ch - ''; ch = getchar();}
   if(last == '-') ans = -ans;
   return ans;
 }
 inline void write(ll x)
 {
   if(x < ) x = -x, putchar('-');
   if(x >= ) write(x / );
   putchar(x %  + '');
 }
 
 int n, a[maxn], pos[maxn];
 int Max[maxn];
 
 int c[maxn];
 int lowbit(int x)
 {
   return x & -x;
 }
 void add(int pos, int x)
 {
   for(; pos <= n && c[pos] < x; pos += lowbit(pos)) c[pos] = x;
 }
 int query(int pos)
 {
   int ret = ;
   for(; pos; pos -= lowbit(pos)) if(c[pos] > ret) ret = c[pos];
   return ret;
 }
 
 int main()
 {
   n = read();
   for(int i = ; i <= n; ++i) a[i] = read();
   for(int i = ; i <= n; ++i) {int x = read(); pos[x] = i;}
   for(int i = ; i <= n; ++i)
     {
       for(int j = max(, a[i] - ); j <= min(n, a[i] + ); ++j)
     Max[j] = query(pos[j] - );
       for(int j = max(, a[i] - ); j <= min(n, a[i] + ); ++j)
     add(pos[j], Max[j] + );
     }
   write(query(n)), enter;
   return ;
 }