BubbleCup - Codeforces 1045C

标签：any   register   mes   code   while   +=   ide   char   har   

Heavy Light Decomposition & Tarjan

Description

Given a simple connected graph.

It is guaranteed that any set of points in a simple cycle includes a complete subgraph.

Query Q times for the shortest path between two given points.

Solution

Every complete graph is a Biconnected Component of the graph.

Thus we can construct a block-cut tree of this graph and calculate the path between points.

Because of the extra distance between cut point and newly biconnected component point, we need to divide result by 2.

//maki
#pragma GCC optimize(2)
#include <bits/stdc++.h>

using namespace std;

int n, m, q;

inline int rd()
{
    register int x = 0; register char c = getchar();
    while(!isdigit(c)) c = getchar();
    while(isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();
    return x;
}

namespace DCC
{
    const int maxn = 1e+5 + 5;
    const int maxm = 5e+5 + 5;
    struct edge
    {
        int to, nxt;
    }e[maxm << 1];
    struct cedge
    {
        int to, nxt;     
    }ce[maxn << 2];
    int dfn[maxn], low[maxn], cnt, ID;
    int lnk[maxn], clnk[maxn << 1], cptr, ptr;
    stack<int> st;

    void add(int bgn, int end)
    {
        e[++ptr].to = end;
        e[ptr].nxt = lnk[bgn];
        lnk[bgn] = ptr;
    }
    void cadd(int bgn, int end)
    {
        ce[++cptr].to = end;
        ce[cptr].nxt = clnk[bgn];
        clnk[bgn] = cptr;
    }

    void tarjan(int x, int inv)
    {
        dfn[x] = low[x] = ++cnt;
        st.push(x);
        for(int p = lnk[x]; p; p = e[p].nxt)
        {
            int y = e[p].to;
            if(!dfn[y])
            {
                tarjan(y, x);
                low[x] = min(low[x], low[y]);
                if(low[y] >= dfn[x]){
                    ID++; int now;
                    cadd(x, ID);
                    do {
                        now = st.top(); st.pop();
                        cadd(ID, now);
                    } while(now != y);
                }
            }
            else if(y != inv) low[x] = min(low[x], dfn[y]);
        }
    }

    void main()
    {
        ID = n;
        tarjan(1, 0);
        //cerr << "tarjan" << endl;
        return;
    }
};
namespace LHD
{
    using namespace DCC;
    const int maxp = 2e+5 + 5;
    int DFN[maxp], siz[maxp], mxson[maxp], dep[maxp];
    int f[maxp], top[maxp];
    //int dis[maxp];

    void dfs1(int x, int fa, int d)
    {
        //cerr << "dfs1" << endl;
        f[x] = fa;
        siz[x] = 1;
        dep[x] = d;
        for(register int p = clnk[x]; p; p = ce[p].nxt)
        {
            int y = ce[p].to;
            if(y == fa) continue;
            dfs1(y, x, d + 1);
            siz[x] += siz[y];
            if(siz[y] > siz[mxson[x]]) mxson[x] = y;
        }
    }
    void dfs2(int x, int init)
    {
        //cerr << "dfs2" << endl;
        top[x] = init;
        if(!mxson[x]) return;
        dfs2(mxson[x], init);
        for(register int p = clnk[x]; p; p = ce[p].nxt)
        {
            int y = ce[p].to;
            if(y == f[x] || y == mxson[x]) continue;
            dfs2(y, y);
        }
    }
    int lca(int x, int y)
    {
        while(top[x] != top[y])
        {
            if(dep[top[x]] < dep[top[y]]) swap(x, y);
            x = f[top[x]];
        }
        return (dep[x] < dep[y]) ? x : y;
    }
    void main()
    {
        dfs1(1, 0, 1);
        dfs2(1, 1);
        //cerr << "dfs" << endl;
        int x, y;
        while(q--)
        {
            //cin >> x >> y;
            x = rd(); y = rd();
            int LCA = lca(x, y);
            int dis = dep[x] + dep[y] - 2 * dep[LCA];
            cout << (dis / 2) << endl;
        }
    }
};

int main()
{
    
    //cin >> n >> m >> q;
    n = rd(); m = rd(); q = rd();
    int u, v;
    for(register int i = 1; i <= m; ++i)
    {
        //cin >> u >> v;
        u = rd(); v = rd();
        DCC::add(u, v); DCC::add(v, u);
    }
    DCC::main();
    LHD::main();
    return 0;
}

BubbleCup - Codeforces 1045C

标签：any   register   mes   code   while   +=   ide   char   har   

原文地址：https://www.cnblogs.com/nishikino-curtis/p/9695155.html