UVA-12363 Hedge Mazes

Hedge Mazes

判断两点之间是不是只有唯一一条简单路径

tarjan 割边

如果两点之间仅有一条简单路径，说明如果将路径上的一条边割掉，则两点之间不连通，换句话说，这条最短路径上的所有边都是割边

因此考虑找出所有的割边，用割边来跑一次并查集

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 2e5 + 10;
int head[maxn], nexx[maxn], to[maxn], vp = 0;
int dfn[maxn], low[maxn], top[maxn], tp = 0;
int cut[maxn];

void init(int n)
{
    for(int i=0; i<=n; i++) head[i] = dfn[i] = low[i] = 0;
    for(int i=0; i<=n; i++) top[i] = i;
    for(int i=0; i<=vp; i++) cut[i] = 0;
    vp = 1;
}

inline void add(int u, int v)
{
    vp++;
    nexx[vp] = head[u];
    head[u] = vp;
    to[vp] = v;
}

int query(int x)
{
    return x == top[x] ? x : top[x] = query(top[x]);
}

void tarjan(int now, int pre)
{
    dfn[now] = low[now] = ++tp;
    for(int i=head[now]; i; i=nexx[i])
    {
        int nex = to[i];
        if(dfn[nex] == 0) 
        {
            tarjan(nex, i);
            low[now] = min(low[nex], low[now]);
            if(low[nex] > dfn[now])
                cut[i] = 1;
        }
        else if(i != (pre ^ 1))
            low[now] = min(low[now], dfn[nex]);
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, m, q;
    while(cin >> n >> m >> q && (n | m | q))
    {
        init(n);
        while(m--)
        {
            int a, b;
            cin >> a >> b;
            add(a, b);
            add(b, a);
        }
        for(int i=1; i<=n; i++)
            if(dfn[i] == 0) tarjan(i, -1);
        for(int i=1; i<=n; i++)
        {
            for(int j=head[i]; j; j=nexx[j])
            {
                if(cut[j] == 0) continue;
                int nex = to[j];
                int a = query(i), b = query(nex);
                if(a != b) top[a] = top[b];
            }
        }
        while(q--)
        {
            int a, b;
            cin >> a >> b;
            int aa = query(a), bb = query(b);
            if(top[aa] == top[bb]) cout << "Y\n";
            else cout << "N\n";
        }
        cout << "-\n";
    }
    // cout << endl;
    return 0;
}