2022杭电多校第三场 L题 Two Permutations

题目链接

Two Permutations

题目大意

给定两个长度为$n$的排列$a,b$和一个由两个排列组成的长度为$2\ast n$的数组$c$，每次从两个排列中任意一个的最左端取出一个数放到$c[i]$里，请问构造出$c$数组的方案数。

题解

比赛的时候想了线性DP的做法，感觉很牛逼。后来看题解，发现大家的做法更优秀。。。尤其是jiangly的模拟大法，让我大受震撼！！！
设$f[i][j][k]$ 表示第$i$个数从$j$数组中选，$i-1$个数从$k$数组中选，还需要一个额外的数组$g$记录选择哪个数组。具体转移看代码~ $(0<=j<=1)(0<=k<=1)$

代码

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define IOS ios::sync_with_stdio(false); cin.tie(0);cout.tie(0);
#define x first
#define y second
#define endl '\n' 
const int inf = 1e9 + 10;
const int maxn = 300010, M = 2000010;
const int mod = 998244353;
typedef pair<int,int> PII;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
int h[maxn], e[M], w[M], ne[M], idx;
int dx[4] = {
    -1, 0, 1, 0}, dy[4] = {
    0, -1, 0, 1};
int cnt;
void add(int a, int b, int c)
{
    
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++;
}
void add(int a, int b)
{
    
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}
int n;
int a[maxn], b[maxn], c[maxn * 2];
int f[maxn * 2][2][2];
int g[maxn * 2][2][2];
int main()
{
    
    int t; scanf("%d", &t);
    while(t --)
    {
    
        scanf("%d", &n);
        for(int i = 1; i <= n; i ++) scanf("%d", &a[i]);
        for(int i = 1; i <= n; i ++) scanf("%d", &b[i]);
        for(int i = 1; i <= n * 2; i ++) scanf("%d", &c[i]);
        memset(f, 0, sizeof f); memset(g, 0, sizeof g);
        if(c[1] == a[1] && c[2] == a[2]) {
    
            f[2][0][0] = 1;
            g[2][0][0] = 2;
        }
        if(c[1] == a[1] && c[2] == b[1]){
    
            f[2][1][0] = 1;
            g[2][1][0] = 1;
        }
        if(c[1] == b[1] && c[2] == a[1]){
    
            f[2][0][1] = 1;
            g[2][0][1] = 1;
        }
        if(c[1] == b[1] && c[2] == b[2]){
    
            f[2][1][1] = 1;
            g[2][1][1] = 0;
        }
        for(int i = 3; i <= n * 2; i ++){
    
            // f[i][0][0]
            int id0 = g[i - 1][0][0], id1 = g[i - 1][0][1];
         // cout << id0 << ' ' << id1 << endl;
            if(a[id0 + 1] == c[i]) f[i][0][0] = (f[i][0][0] + f[i - 1][0][0]) % mod, g[i][0][0] = id0 + 1;
            if(a[id1 + 1] == c[i]) f[i][0][0] = (f[i][0][0] + f[i - 1][0][1]) % mod, g[i][0][0] = id1 + 1;

            // f[i][0][1]
            id0 = g[i - 1][1][0], id1 = g[i - 1][1][1];
            if(a[id0 + 1] == c[i]) f[i][0][1] = (f[i][0][1] + f[i - 1][1][0]) % mod, g[i][0][1] = id0 + 1;
            if(a[id1 + 1] == c[i]) f[i][0][1] = (f[i][0][1] + f[i - 1][1][1]) % mod, g[i][0][1] = id1 + 1;

            // f[i][1][0]
            id0 = g[i - 1][0][0], id1 = g[i - 1][0][1];
            if(b[(i - id0)] == c[i]) f[i][1][0] = (f[i][1][0] + f[i - 1][0][0]) % mod, g[i][1][0] = id0;
            if(b[i - id1] == c[i]) f[i][1][0] = (f[i][1][0] + f[i - 1][0][1]) % mod, g[i][1][0] = id1;

            // f[i][1][1]
            id0 = g[i - 1][1][0], id1 = g[i - 1][1][1];
            if(b[i - id0] == c[i])  f[i][1][1] = (f[i][1][1] + f[i - 1][1][0]) % mod, g[i][1][1] = id0;
            if(b[i - id1] == c[i]) f[i][1][1] = (f[i][1][1] + f[i - 1][1][1]) % mod, g[i][1][1] = id1;
        }
        int res = 0;
        for(int i = 0; i < 2; i ++){
    
            for(int j = 0; j < 2; j ++){
    
                res = (res + f[n * 2][i][j]) % mod;
            }
        }
        printf("%d\n", res);
    }
    return 0;
}