• [CF786B] Legacy


    [CF786B] Legacy - 线段树优化建图,最短路

    Description

    有三种操作:1.进行单点与单点连有向边 2.进行单点与区间连有向边 3.进行区间与单点连有向边。求最短路。

    Solution

    发现没有区间和区间,这暗示的够明显了,线段树优化建图即可

    #include <bits/stdc++.h>
    using namespace std;
    
    #define int long long
    
    #ifndef _SP_H_
    #define _SP_H_
    
    #include <bits/stdc++.h>
    using namespace std;
    #define reset(x) memset(x, 0, sizeof x)
    #define reset3f(x) memset(x, 0x3f, sizeof x)
    namespace sp
    {
        const int N = 2e+6 + 5;
        vector<pair<int, int>> g[N];
        int n, v0 = 1, d[N], v[N];
        void make(int t1, int t2, int t3)
        {
            g[t1].push_back(make_pair(t2, t3));
        }
        void reset_graph()
        {
            for (int i = 0; i <= n; i++)
                g[i].clear();
        }
        void solve()
        {
            priority_queue<pair<int, int>> qu;
            reset3f(d);
            reset(v);
            d[v0] = 0;
            qu.push(make_pair(0, v0));
            while (qu.size())
            {
                int p = qu.top().second, r = qu.top().first;
                qu.pop();
                if (r + d[p])
                    continue;
                for (int i = 0; i < g[p].size(); i++)
                {
                    int q = g[p][i].first, w = g[p][i].second;
                    if (d[q] > d[p] + w)
                    {
                        d[q] = d[p] + w;
                        qu.push(make_pair(-d[q], q));
                    }
                }
            }
        }
    } // namespace sp
    #endif
    
    int n;
    
    void solvesp(int s)
    {
        sp::solve();
        sp::v0 = s;
        sp::n = n * 12;
        sp::solve();
        for (int i = 1; i <= n; i++)
            if (sp::d[i] <= 1e15)
                cout << sp::d[i] << " ";
            else
                cout << -1 << " ";
    }
    
    void make(int u, int v, int w)
    {
        sp::make(u, v, w);
    }
    
    void build(int p, int l, int r)
    {
        if (l == r)
        {
            int i = l;
            make(p + 4 * n, i, 0);
            make(i, p + 8 * n, 0);
        }
        else
        {
            build(p * 2, l, (l + r) / 2);
            build(p * 2 + 1, (l + r) / 2 + 1, r);
            make(p + 4 * n, p * 2 + 4 * n, 0);
            make(p + 4 * n, p * 2 + 1 + 4 * n, 0);
            make(p * 2 + 4 * n + 4 * n, p + 4 * n + 4 * n, 0);
            make(p * 2 + 1 + 4 * n + 4 * n, p + 4 * n + 4 * n, 0);
        }
    }
    
    void query(int p, int l, int r, int ql, int qr, vector<int> &ans)
    {
        if (l > qr || r < ql)
            return;
        if (l >= ql && r <= qr)
        {
            ans.push_back(p);
        }
        else
        {
            query(p * 2, l, (l + r) / 2, ql, qr, ans);
            query(p * 2 + 1, (l + r) / 2 + 1, r, ql, qr, ans);
        }
    }
    
    vector<int> getset(int l, int r)
    {
        vector<int> ans;
        query(1, 1, n, l, r, ans);
        return ans;
    }
    
    signed main()
    {
        ios::sync_with_stdio(false);
    
        int q, s;
        cin >> n >> q >> s;
        build(1, 1, n);
        for (int i = 1; i <= q; i++)
        {
            int t;
            cin >> t;
            if (t == 1)
            {
                int u, v, w;
                cin >> u >> v >> w;
                make(u, v, w);
            }
            else if (t == 2)
            {
                int u, l, r, w;
                cin >> u >> l >> r >> w;
                auto vec = getset(l, r);
                for (auto i : vec)
                    make(u, i + 4 * n, w);
            }
            else if (t == 3)
            {
                int u, l, r, w;
                cin >> u >> l >> r >> w;
                auto vec = getset(l, r);
                for (auto i : vec)
                    make(i + 8 * n, u, w);
            }
        }
        solvesp(s);
    }
    
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  • 原文地址:https://www.cnblogs.com/mollnn/p/14361264.html
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