Wannafly挑戰賽2_D Delete(拓撲序+最短路+線段樹)

Wannafly挑戰賽2_D Delete

Problem :
給定一張n個點，m條邊的帶權有向無環圖，同時給定起點S和終點T，一共有q個詢問，每次詢問刪掉某個點和全部與它相連的邊以後S到T的最短路，詢問之間互相獨立(即刪除操做在詢問結束以後會當即撤銷)，若是刪了那個點後不存在S到T的最短路，則輸出-1。
n,q <= 10^5
Solution :
注意到題中所給的是DAG，首先能夠找出圖中結點的拓撲序。對於刪除掉某個點以後，若仍存在一條從S到T的最短路，那麼對應到拓撲序中，必然有一條邊跨過了該點(即這條邊的兩個端點的拓撲序在這個點的兩邊)。故對於每條邊(u, v, w)， 對拓撲序(a[u], a[v])中的點提供了一條長度爲ds[u]+dt[v]+w的最短路，用線段樹來維護最小值。
注意特判若u->v不可能通過這個點，那麼直接輸出最短路。

#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();
}