题目链接

Codeforces Round 883 (Div. 3) G. Rudolf and CodeVid-23

思路

因为$n$最大值为$10$，且只有$01$两种状态，当作二进制数转化为十进制数后最多只有$1024$种。

因为$m$的最大值为$1e3$，因此我们可以枚举每一种状态（$2^n$种）和每一种药（$m$种），通过药的作用和副作用对所有的状态进行连边建图。

最后，跑Dijkstra求初始状态到$0$号点的最短路即可。

代码

#pragma GCC optimize("O2")
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int, int>pii;
const int N = 2e3 + 5, M = 1e6 + 5;
const int mod = 998244353;
const int inf = 0x3f3f3f3f3f3f3f3f;int n, m, maxx;
int d[N], dist[N];
bool st[N];
string s, s1[N], s2[N];
int mp[N][N];
void dijkstra()
{int start = 0;for (int i = 0; i < n; i++){if (s[i] == '1')start += pow(2, n - i - 1);}dist[start] = 0;priority_queue<pii, vector<pii>, greater<pii>>q;q.push({0, start});while (q.size()){int u = q.top().second;int w = q.top().first;q.pop();if (st[u]) continue;st[u] = true;for (int v = 0; v <= maxx; v++){if (u == v) continue;int nw = mp[u][v];if (nw == inf) continue;if (dist[v] > w + nw){dist[v] = w + nw;q.push({dist[v], v});}}}
}
void solve()
{cin >> n >> m;cin >> s;maxx = pow(2, n) - 1;for (int i = 0; i <= maxx; i++){dist[i] = inf;st[i] = false;for (int j = 0; j <= maxx; j++)mp[i][j] = inf;}for (int i = 1; i <= m; i++){cin >> d[i] >> s1[i] >> s2[i];}for (int i = 1; i <= m; i++){for (int j = 0; j <= maxx; j++){string v = "";for (int k = n - 1; k >= 0; k--){int bit = (j >> k) & 1;if (bit) v.push_back('1');else v.push_back('0');}for (int k = 0; k < n; k++){if (s1[i][k] == '1')v[k] = '0';}for (int k = 0; k < n; k++){if (s2[i][k] == '1')v[k] = '1';}int res = 0;reverse(v.begin(), v.end());for (int k = 0; k < n; k++){if (v[k] == '1')res += pow(2, k);}if (j == res) continue;mp[j][res] = min(mp[j][res], d[i]);}}dijkstra();if (dist[0] == inf){cout << -1 << endl;}else cout << dist[0] << endl;
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int test = 1;cin >> test;for (int i = 1; i <= test; i++){solve();}return 0;
}