2019 Multi-University Training Contest 6

暴力LIS

反着做比较好写，把解冻看成冻住，每次看冻住的位置是不是当前LIS中的位置，如果不是就直接删除，如果是，就暴力重新计算一次LIS

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define __fastIn ios::sync_with_stdio(false), cin.tie(0)
#define pb push_back
using namespace std;
using LL = long long;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        ret = (ret << 3) + (ret << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -ret : ret;
}
template <typename A>
inline A __lcm(A a, A b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 50005;
int _, n, a[N], fr[N], dp[N], f[N], lis, ans[N];
bool vis[N], pos[N];

void calc(){
    full(f, INF), full(dp, 0), lis = 0;
    for(int i = 1; i <= n; i ++){
        if(!vis[i]) continue;
        int p = (int)(lower_bound(f + 1, f + n + 1, a[i]) - f);
        if(f[p] == INF){
            f[p] = a[i], dp[i] = p;
            lis = max(lis, p);
        }
        else{
            f[p] = a[i], dp[i] = p;
        }
    }
    full(pos, false);
    int len = lis;
    for(int i = n; i >= 1; i --){
        if(!vis[i]) continue;
        if(!lis) break;
        if(dp[i] == len){
            pos[i] = true, len --;
        }
    }
}

int main(){

    for(_ = read(); _; _ --){
        n = read();
        for(int i = 1; i <= n; i ++) a[i] = read();
        for(int i = 1; i <= n; i ++) fr[i] = read();
        full(vis, true), full(ans, 0);
        calc();
        for(int i = n; i >= 1; i --){
            ans[i] = lis;
            vis[fr[i]] = false;
            if(pos[fr[i]]) calc();
        }
        for(int i = 1; i <= n; i ++)
            printf("%d%c", ans[i], " 
"[i == n]);
    }
    return 0;
}