POI2007ATR-Tourist Attractions

最短路 #状压dp #滚动优化 #POI #Year2007

从前 \(k\) 个跑 \(dijksta\) ，对这 \(k\) 个点到达的状态状压

会 MLE ，考虑每次转移都只会增加一个状压下的 \(1\) ，按照 \(popcount\) 分组做滚动

// Author: xiaruize
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

int n, m, k;
vector<pii> g[50050];
int dis[25][50050];
int st[50050];
int cnt[1050000], id[1050000];
vector<int> vec[23];
int dp[190000][2][23];

void dijkstra(int x)
{
	mms(dis[x], 0x3f);
	dis[x][x] = 0;
	priority_queue<pii, vector<pii>, greater<pii>> q;
	q.push({0, x});
	while (!q.empty())
	{
		auto [ds, u] = q.top();
		q.pop();
		if (ds != dis[x][u])
			continue;
		for (auto [v, w] : g[u])
		{
			if (dis[x][v] > dis[x][u] + w)
			{
				dis[x][v] = dis[x][u] + w;
				q.push({dis[x][v], v});
			}
		}
	}
}

void solve()
{
	cin >> n >> m >> k;
	k++;
	rep(i, 1, m)
	{
		int u, v, w;
		cin >> u >> v >> w;
		g[u].push_back({v, w});
		g[v].push_back({u, w});
	}
	rep(i, 1, k) dijkstra(i);
	cin >> m;
	rep(i, 1, m)
	{
		int u, v;
		cin >> u >> v;
		st[v] |= (1 << u - 2);
	}
	rep(i, 2, k) st[i] |= (1 << i - 2);
	if (k == 1)
	{
		cout << dis[1][n] << endl;
		return;
	}
	rep(msk, 1, (1 << k - 1) - 1)
	{
		int t = __builtin_popcount(msk);
		vec[t].push_back(msk);
		cnt[t]++;
		id[msk] = cnt[t];
	}
	int res = INF;
	rep(i, 1, k - 1)
	{
		for (auto v : vec[i])
		{
			mms(dp[id[v]][i & 1], 0x3f);
			rep(cur, 2, k)
			{
				if ((st[cur] | v) != v)
					continue;
				if (v - (v & -v) != 0)
				{
					rep(x, 2, k)
					{
						if (x != cur && (v & (1 << x - 2)))
							dp[id[v]][i & 1][cur] = min(dp[id[v]][i & 1][cur], dp[id[v ^ (1 << cur - 2)]][(i & 1) ^ 1][x] + dis[x][cur]);
					}
				}
				else
					dp[id[v]][i & 1][cur] = dis[1][cur];
				if (v == (1 << k - 1) - 1)
				{
					res = min(res, dp[id[v]][i & 1][cur] + dis[cur][n]);
				}
			}
		}
	}
	cout << res << endl;
}

#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif

signed main()
{
	// freopen(".in","r",stdin);
	// freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	// cin >> testcase;
	while (testcase--)
		solve();
#ifndef ONLINE_JUDGE
	cerr << "Memory use:" << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB" << endl;
	cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms" << endl;
#endif
	return 0;
}