我是记 $f_{i,j,k,l}$ 表示处理到第 $i$ 个箱子，所有钥匙的残余流量为 $j$，选到了第 $k$ 把钥匙，箱子残余流量为 $l$ 的最小代价。

看了半天没看出错误，球球大佬帮我看看嘛

#include<bits/stdc++.h>
#define int long long
#define N 10
using namespace std;
int n, m, ans = 1e18;
int a[N], b[N], c[N][N], cnt;
vector<int>now, z, p[100005], r;
int f[8][16000][7][7];
map<vector<int>, int>num;
void dfs(int k) {
	if(k > m) {
		//for(auto x : z) cout << x << " ";
		//cout << endl;
		p[++cnt] = z;
		num[z] = cnt;
		for(int i = 1; i <= n + 1; i++)
		for(int j = 1; j <= m; j++)
		for(int l = 0; l <= 4; l++)
			f[i][cnt][j][l] = 1e18;
		return;
	}
	for(int i = 0; i <= b[k]; i++) {
		z.push_back(i);
		dfs(k + 1);
		z.pop_back();
	}
}
void upd(int &x, int y) {
	x = min(x, y);
}
signed main() {
	//ios::sync_with_stdio(false);
	//cin.tie(0); cout.tie(0);
	cin >> n >> m;
	for(int i = 1; i <= n; i++) cin >> a[i];
	for(int i = 1; i <= m; i++) cin >> b[i], now.push_back(b[i]);
	for(int i = 1; i <= n; i++)
	for(int j = 1; j <= m; j++) cin >> c[i][j];
	dfs(1);
	//cout << cnt << endl;
	//cout << num[now] << endl;
	f[1][num[now]][1][a[1]] = 0;
	for(int i = 1; i <= n; i++) {
		for(int k = 1; k <= m; k++) {
			for(int j = 1; j <= cnt; j++) {
				for(int l = 0; l <= a[i]; l++) {
					if(k != m) {
						upd(f[i][j][k + 1][l], f[i][j][k][l]);
						int use = min(l, p[j][k - 1]);
						r = p[j];
						r[k - 1] -= use;
						upd(f[i][num[r]][k + 1][l - use], f[i][j][k][l] + c[i][k]);
					}
					else {
						if(l == 0) upd(f[i + 1][j][1][a[i + 1]], f[i][j][k][l]);
						else if(p[j][k - 1] >= l) {
							r = p[j];
							r[k - 1] -= l;
							upd(f[i + 1][num[r]][1][a[i + 1]], f[i][j][k][l] + c[i][k]);
						}
					}
				}
			}
		}
	}
	/*for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= cnt; j++) {
			for(int k = 1; k <= m; k++) {
				for(int l = 0; l <= a[i]; l++) {
					if(f[i][j][k][l] == 1e18) continue;
					printf("f[%lld][%lld][%lld][%lld] = %lld\n", i, j, k, l, f[i][j][k][l]);
				}
			}
		}
	}*/
	for(int i = 1; i <= cnt; i++) ans = min(ans, f[n + 1][i][1][0]);
	if(ans == 1e18) cout << -1;
	else cout << ans;
	return 0;
}
/*
2 3
3 3
1 1 4
10 20 100
20 15 80
*/