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 = + ;
int a[maxn], b[maxn], c[maxn * ];
int vis[maxn][maxn], DFN, dp[maxn][maxn];
bool dfs(int cur, int pos1, int pos2) {
if (cur == n1 + n2 + ) return true;
if (vis[pos1][pos2] == DFN) return dp[pos1][pos2];
vis[pos1][pos2] = DFN;
if (a[pos1] == c[cur]) {
if (dfs(cur + , pos1 + , pos2)) {
dp[pos1][pos2] = true;
return true;
}
}
if (b[pos2] == c[cur]) {
if (dfs(cur + , pos1, pos2 + )) {
dp[pos1][pos2] = true;
return true;
}
}
dp[pos1][pos2] = false;
return false;
}
void work() {
++DFN;
for (int i = ; i <= n1; ++i) scanf("%d", &a[i]);
for (int i = ; i <= n2; ++i) scanf("%d", &b[i]);
a[n1 + ] = b[n2 + ] = inf;
for (int i = ; i <= n1 + n2; ++i) scanf("%d", &c[i]);
if (dfs(, , )) {
printf("possible\n");
} else printf("not possible\n");
} 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 ;
}