Wannafly挑战赛2_D Delete(拓扑序+最短路+线段树)

Wannafly挑战赛2_D Delete

Problem :

给定一张n个点，m条边的带权有向无环图，同时给定起点S和终点T，一共有q个询问，每次询问删掉某个点和所有与它相连的边之后S到T的最短路，询问之间互相独立(即删除操作在询问结束之后会立即撤销)，如果删了那个点后不存在S到T的最短路，则输出-1。

n,q <= 10^5

Solution :

注意到题中所给的是DAG，首先可以找出图中结点的拓扑序。对于删除掉某个点之后，若仍存在一条从S到T的最短路，那么对应到拓扑序中，必然有一条边跨过了该点(即这条边的两个端点的拓扑序在这个点的两边)。故对于每条边(u, v, w)， 对拓扑序(a[u], a[v])中的点提供了一条长度为ds[u]+dt[v]+w的最短路，用线段树来维护最小值。

注意特判若u->v不可能经过这个点，那么直接输出最短路。

#include <bits/stdc++.h>
using namespace std;

#define endl "\n"
const long long INF = 1ll << 60;
const int N = 1e5 + 8;
struct edge
{
	int v, w;
	edge(int v = 0, int w = 0):v(v),w(w){}
};
vector <edge> vec[N], vec2[N];
long long ds[N], dt[N];
int a[N], deg[N];
int n, m, S, T;

void init()
{
	cin >> n >> m >> S >> T;
	for (int i = 1; i <= n; ++i) vec[i].clear(), vec2[i].clear();
	for (int i = 1; i <= m; ++i)
	{
		int u, v, w; cin >> u >> v >> w;
		vec[u].push_back(edge(v, w));
		vec2[v].push_back(edge(u, w));
		deg[v]++;
	}
}
void getDag()
{
	static queue <int> Q;
	int tot = 0;
	for (int i = 1; i <= n; ++i)
	{
		if (deg[i] == 0)
		{
			a[i] = ++tot;
			Q.push(i);
		}
	}
	while (!Q.empty())
	{
		int u = Q.front(); Q.pop();
		for (auto p : vec[u])
		{
			deg[p.v]--;
			if (deg[p.v] == 0)
			{
				a[p.v] = ++tot;
				Q.push(p.v);
			}
		}
	}
}
struct node
{
	int u;
	long long w;
	bool operator < (const node &b) const
	{
		return w > b.w;
	}
};
void Dijkstra(int S, long long dis[], vector <edge> vec[])
{
	static priority_queue<node> Q;
	static bool vis[N];
	for (int i = 1; i <= n; ++i) dis[i] = INF, vis[i] = 0;
	Q.push((node){S, 0});
	dis[S] = 0;
	while (!Q.empty())
	{
		int u = Q.top().u; Q.pop();
		if (vis[u]) continue;
		vis[u] = 1;
		for (auto p : vec[u])
			if (dis[u] + p.w < dis[p.v])
			{
				dis[p.v] = dis[u] + p.w;
				Q.push((node{p.v, dis[p.v]}));
			}
	}
}
class SegmentTree
{
public:
	long long tag[N << 2];
	void build(int l, int r, int rt)
	{
		tag[rt] = INF;
		if (l == r) return;
		int m = l + r >> 1;
		build(l, m, rt << 1);
		build(m + 1, r, rt << 1 | 1);
	}
	void update(int L, int R, long long val, int l, int r, int rt)
	{
	//	if (l == 1 && r == n) cout << L << " " << R << " " << val << endl;
		if (L <= l && r <= R)
		{
			tag[rt] = min(tag[rt], val);
			return;
		}
		int m = l + r >> 1;
		if (L <= m) update(L, R, val, l, m, rt << 1);
		if (m <  R) update(L, R, val, m + 1, r, rt << 1 | 1);
	}
	void query(int x, int l, int r, int rt, long long &ans)
	{
		ans = min(ans, tag[rt]);
		if (l == r) return;
		int m = l + r >> 1;
		if (x <= m) query(x, l, m, rt << 1, ans);
		else query(x, m + 1, r, rt << 1 | 1, ans);
	}
}ST;
void buildSeg()
{
	ST.build(1, n, 1);
	for (int u = 1; u <= n; ++u)
		for (auto p : vec[u])
			if (a[u] + 1 < a[p.v] && ds[u] != INF && dt[p.v] != INF)
		{
		//	cout << u << " " << p.v << " " << a[u] << " " << a[p.v] << endl;
			ST.update(a[u] + 1, a[p.v] - 1, ds[u] + dt[p.v] + p.w, 1, n, 1);
		}
}
void solve()
{
	int Q; cin >> Q;
	for (; Q; --Q)
	{
		int u; cin >> u;
		long long ans = INF;
		if (ds[u] == INF || dt[u] == INF)
		{
			cout << dt[S] << endl;
			continue;
		}
		ST.query(a[u], 1, n, 1, ans);
		if (ans == INF) cout << -1 << endl; else cout << ans << endl;
	}
}
int main()
{
	cin.sync_with_stdio(0);
	init();
	getDag();
	Dijkstra(S, ds, vec);
	Dijkstra(T, dt, vec2);
	buildSeg();
	solve();
}