Railroad UVALive

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2889

做这题的时候我第一感觉是直接dfs的dp，用dp[i][j]表示第一个数组处理到第i位数字，第二个数组处理到第j个数字，能否组合出第三个数组的前cur个数字。

转移的话，就是试试第i为能否匹配第cur个了，如果可以就直接搜，不可以就看看第j位，都不可以就false

这样转移起来我觉得很明朗，实在想不到怎么用数组来dp。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
int n1, n2;
const int maxn = 1000 + 20;
int a[maxn], b[maxn], c[maxn * 2];
int vis[maxn][maxn], DFN, dp[maxn][maxn];
bool dfs(int cur, int pos1, int pos2) {
    if (cur == n1 + n2 + 1) return true;
    if (vis[pos1][pos2] == DFN) return dp[pos1][pos2];
    vis[pos1][pos2] = DFN;
    if (a[pos1] == c[cur]) {
        if (dfs(cur + 1, pos1 + 1, pos2)) {
            dp[pos1][pos2] = true;
            return true;
        }
    }
    if (b[pos2] == c[cur]) {
        if (dfs(cur + 1, pos1, pos2 + 1)) {
            dp[pos1][pos2] = true;
            return true;
        }
    }
    dp[pos1][pos2] = false;
    return false;
}
void work() {
    ++DFN;
    for (int i = 1; i <= n1; ++i) scanf("%d", &a[i]);
    for (int i = 1; i <= n2; ++i) scanf("%d", &b[i]);
    a[n1 + 1] = b[n2 + 1] = inf;
    for (int i = 1; i <= n1 + n2; ++i) scanf("%d", &c[i]);
    if (dfs(1, 1, 1)) {
        printf("possible
");
    } else printf("not possible
");
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    while (scanf("%d%d", &n1, &n2) != EOF && (n1 + n2)) work();
    return 0;
}