CF555E Case of Computer Network

传送门

题意

题解

容易发现，在同一个边双内的点我们是不用管它的，一定可以。

那么我们缩点，然后把图变成一颗树。这样，$s$ 到 $t$ 的路径就可以用$\mathrm{LCA}$求了。

我们把 $s$ 到 $\mathrm{LCA}$ 的路径打向上的标记，把 $\mathrm{LCA}$ 到 $t$ 的路径打向下的标记。

只要没有路径有两种标记，就是可以的。

打标记使用树上差分实现。时间复杂度：$\mathcal O\left(n+m+q\mathrm{lg}n\right)$。

附上代码：

#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;

#define down(x, y) x = min(x, y)

int n, m, q;
struct edge {
    int from, to, nxt, id;
} edges[400005];
struct edgE {
    int to, nxt;
} edgEs[400005];
int e0, e = 1, e2 = 1, hed[200005], hEd[200005], tim = 0, fa[200005], dfn[200005], low[200005], scc[200005], dis[200005], lca[18][200005];
int flg[200005]; bool root[200005]; bool goon[200005];
int leaves[200005], top=0;
const int CLEAN_FLAG = 0x1;
const int UP_FLAG = 0x2;
const int DOWN_FLAG = 0x4;
stack<int> S;
inline bool hasflag(int x, int attr) {
    return flg[x] & attr;
}
inline bool setflag(int x, int attr) {
    if (attr == UP_FLAG && hasflag(x, DOWN_FLAG)) return false;
    if (attr == DOWN_FLAG && hasflag(x, UP_FLAG)) return false;
    flg[x] |= attr;
    return true;
}
inline void addedge(int x, int y) {
    edges[e] = (edge){x, y, hed[x], e2};
    hed[x] = e++;
    edges[e] = (edge){y, x, hed[y], e2++};
    hed[y] = e++;
}
inline void addedgE(int x, int y) {
    edgEs[e] = (edgE){y, hEd[x]};
    hEd[x] = e++;
}
inline void _tarjan(int x) {
    dfn[x] = low[x] = ++tim;
    S.push(x);
    for (int e=hed[x]; e; e=edges[e].nxt) {
        int y = edges[e].to;
        if (edges[e].id == fa[x]) continue;
        if (!dfn[y]) {
            fa[y] = edges[e].id;
            _tarjan(y);
            down(low[x], low[y]);
        } else down(low[x], dfn[y]);
    }
    if (dfn[x] == low[x]) {
        scc[x] = ++e2;
        while (S.top() != x) {
            scc[S.top()] = e2;
            S.pop();
        }
        S.pop();
    }
}
inline void tarjan() {
    for (int i=1; i<=n; i++) if (!dfn[i]) _tarjan(i);
    for (int i = 0; i < e0; i++) {
        int x = edges[i].from, y = edges[i].to;
        if (scc[x] == scc[y]) continue;
        addedgE(scc[x], scc[y]);
    }
}
inline void lca_pre() {
    for (int i=1; i<18; i++)
        for (int j=1; j<=n; j++)
            lca[i][j] = lca[i-1][lca[i-1][j]];
}
inline int LCA(int x, int y) {
    if (dis[x] > dis[y]) swap(x, y);
    int u = dis[x], v = dis[y];
    int X=x, Y=y;
    for (int det = v-u, i=0; det; det>>=1, ++i)
        if (det & 1) Y = lca[i][Y];
    if (X == Y) return X;
    for (int i=17; i>=0; i--) {
        if (lca[i][X] == lca[i][Y]) continue;
        X = lca[i][X]; Y = lca[i][Y];
    }
    return lca[0][X];
}
int group[200005], current;
inline void dfs(int x, int fa) {
    bool isleaf = true; group[x] = current;
    for (int e=hEd[x]; e; e=edgEs[e].nxt) {
        int y = edgEs[e].to;
        if (y != fa) {
            isleaf = false;
            lca[0][y] = x;
            dis[y] = dis[x] + 1;
            dfs(y, x);
        }
    }
    if (isleaf) leaves[top++] = x;
}
inline bool akmachine(int x, int flag) {
    while (!root[x]) {
        if (goon[x]) return true;
        goon[x] = true;
        if (hasflag(x, UP_FLAG) && !hasflag(x, CLEAN_FLAG) && (flag & DOWN_FLAG)
         || hasflag(x, DOWN_FLAG) && !hasflag(x, CLEAN_FLAG) && (flag & UP_FLAG))
            return false;
        if (hasflag(x, CLEAN_FLAG)) flag = 0;
        flag = flg[x] & (UP_FLAG | DOWN_FLAG);
        x = lca[0][x];
    }
    return true;
}
inline bool check() {
    for (int i=0; i<top; i++) if (!akmachine(leaves[i], 0)) return false;
    return true;
}
inline bool solve() {
    for (int i=1; i<=n; i++)
        if (!dis[i]) {
            current = i; lca[0][i] = i; root[i] = 1;
            dfs(i, 0);
        }
    lca_pre();
    while (q--) {
        int s, t; cin >> s >> t;
        s = scc[s]; t = scc[t];
        if (group[s] != group[t]) return false;
        if (s == t) continue;
        int p = LCA(s, t);
        if (p != s) if (!setflag(s, UP_FLAG)) return false;
        if (p != t) if (!setflag(t, DOWN_FLAG)) return false;
        setflag(p, CLEAN_FLAG);
    }
    return check();
}

int main() {
    ios :: sync_with_stdio(false); cin.tie(0);
    cin >> n >> m >> q;
    for (int i=1; i<=m; i++) {
        int x, y; cin >> x >> y;
        addedge(x, y);
    }
    e0 = e; e = 1; e2 = 0; tarjan();
    cout << (solve() ? "Yes" : "No") << endl;
    return 0;
}