tarjan学习笔记

1.$tarjan$求强连通分量

思想：在$dfs$的过程中，把强连通分量中的点入栈，当找到一个强连通分量的最起始的点，就将其所在强连通分量中的点出栈。

缩点

把强连通分量中的点缩成一个点，进行重新建图，从而解决一些问题。

2.割点
若将这个点在图中所连的边删去，图变得不连通，则称这个点为一个割点。

考虑两种情况：

若节点$x$为根节点，则它若联结着两颗及以上数量的子树，则$x$为割点。

$otherwise$，设$x$的其中一个儿子为$v$，若出现$low[v] ge dfn[x]$，则$x$为割点。

3.习题

luogu 3388 【模板】割点（割顶）

#include <bits/stdc++.h>
const int MAXN = 20050;
const int MAXM = 100050;
using namespace std;
struct node {
    int to, nextt;
}edge[MAXM << 1];
int n, m, u, v, num, cnt, low[MAXN], dfn[MAXN], head[MAXN];
set<int> s;
void addedge(int u, int v) {
    edge[++num].to = v;
    edge[num].nextt = head[u];
    head[u] = num;
}
void tarjan(int x, int fa) {
    dfn[x] = low[x] = ++cnt;
    int child = 0;
    for(int i = head[x]; i; i = edge[i].nextt) {
        int to = edge[i].to;
        if(!dfn[to]) {
            if(x == fa)
                child++;
            tarjan(to, x);
            low[x] = min(low[x], low[to]);
            if(x != fa && low[to] >= dfn[x])
                s.insert(x);
        }
        low[x] = min(low[x], dfn[to]);
    }
    if(x == fa && child >= 2)
        s.insert(x);
}
int main() {
    cin >> n >> m;
    for(int i = 1; i <= m; i++) {
        cin >> u >> v;
        addedge(u, v);
        addedge(v, u);
    }
    for(int i = 1; i <= n; i++) {
        if(!dfn[i])
            tarjan(i, i);
    }
    cout << (int)s.size() << endl;
    for(set<int>::iterator it = s.begin(); it != s.end(); it++)
        cout << *it << " ";
    cout << endl;
    return 0;
}

Luogu 2921 Trick or Treat on the Farm

第一种做法是进行$tarjan$，对于所在强连通分量里个数大于$1$的点，答案即为当前强连通分量的大小；由于数据的特殊性，另一种情况即为到强连通分量的距离加上那个强连通分量的大小。

第二种做法是三遍递归。第一遍把所有不在环内的点打上标记，第二遍求出未被标记的点所在环的大小，第三遍累加到环的距离。总体想法与第一种是一样的。

代码(#2)：

#include <bits/stdc++.h>
const int MAXN = 100050;
using namespace std;
int n, x, nextt[MAXN], ans[MAXN], in[MAXN];
bool vis[MAXN];
void dfs1(int x) {
    vis[x] = true;
    if(!--in[nextt[x]])
        dfs1(nextt[x]);
}
int dfs2(int x, int num) {
    vis[x] = true;
    ans[x] = num;
    if(ans[nextt[x]])    
        return num;
    return ans[x] = dfs2(nextt[x], num + 1);
}
int dfs3(int x) {
    if(ans[x])
        return ans[x];
    return ans[x] = dfs3(nextt[x]) + 1;
}
int main() {
    cin >> n;
    for(int i = 1; i <= n; i++) {
        cin >> nextt[i];
        ++in[nextt[i]];
    }
    for(int i = 1; i <= n; i++) {
        if(!vis[i] && !in[i])
            dfs1(i);
    }
    for(int i = 1; i <= n; i++) {
        if(!ans[i] && in[i])
            x = dfs2(i, 1);
    }
    for(int i = 1; i <= n; i++) {
        if(!ans[i] && !in[i])
            x = dfs3(i);
    }
    for(int i = 1; i <= n; i++)
        cout << ans[i] << endl;
    return 0;
}

Luogu 1726 上白泽慧音

显然答案即为最大的强连通分量的个数。

#include <bits/stdc++.h>
const int MAXN = 5050;
const int MAXM = 50050;
const int INF = 1 << 30;
using namespace std;
stack<int> s;
struct Edge {
    int to , nextt;
}edge[MAXM << 1];
int n, m, u, v, ty, num, cnt, cnt1, Max, pos, ss[MAXN], head[MAXN], low[MAXN], dfn[MAXN], sccno[MAXN];
void addedge(int u, int v) {
    edge[++num].to = v;
    edge[num].nextt = head[u];
    head[u] = num;
}
void tarjan(int x) {
    dfn[x] = low[x] = ++cnt;
    s.push(x);
    for(int i = head[x]; i; i = edge[i].nextt) {
        int to = edge[i].to;
        if(!dfn[to]) {
            tarjan(to);
            low[x] = min(low[x], low[to]);
        }
        else if(!sccno[to])
            low[x] = min(low[x], dfn[to]);
    }
    if(dfn[x] == low[x]) {
        int now, cnt2 = 0;
        ++cnt1;
        while(true) {
            now = s.top();
            s.pop();
            ++cnt2;
            sccno[now] = cnt1;
            ss[now] = cnt1;
            if(now == x)
                break;
        }
        if(cnt2 > Max) {
            Max = cnt2;
            pos = cnt1;
        }
    }
}
int main() {
    cin >> n >> m;
    for(int i = 1; i <= m; i++) {
        cin >> u >> v >> ty;
        addedge(u, v);
        if(ty == 2)
            addedge(v, u);
    }
    for(int i = 1; i <= n; i++) {
        if(!dfn[i])
            tarjan(i);
    }
    cout << Max << endl;
    for(int i = 1; i <= n; i++) {
        if(ss[i] == pos) {
            for(int j = i; j <= n; j++) {
                if(ss[j] == pos)
                    cout << j << " ";
            }
            return 0;
        } 
    }
    return 0;
}

Luogu 3387 【模板】缩点

先进行缩点，后作类似拓扑的操作，$dp$即可。

#include <bits/stdc++.h>
const int MAXN = 10050;
const int MAXM = 100050;
using namespace std;
stack<int> ss;
queue<int> q;
struct Edge1 {
    int u, to, nextt;
}edge1[MAXM];
struct Edge2 {
    int u, to, nextt;
}edge2[MAXN];
int n, m, u, v, ans, num1, num2, cnt, cnt1, head1[MAXN], head2[MAXN], dfn[MAXN], low[MAXN], sccno[MAXN], sd[MAXN], in[MAXN], val[MAXN], dis[MAXN];
void addedge1(int u, int v) {
    edge1[++num1].u = u;
    edge1[num1].to = v;
    edge1[num1].nextt = head1[u];
    head1[u] = num1;
}
void addedge2(int u, int v) {
    edge2[++num2].u = u;
    edge2[num2].to = v;
    edge2[num2].nextt = head2[u];
    head2[u] = num2;
}
void tarjan(int x) {
    dfn[x] = low[x] = ++cnt;
    ss.push(x);
    for(int i = head1[x]; i; i = edge1[i].nextt) {
        int to = edge1[i].to;
        if(!dfn[to]) {
            tarjan(to);
            low[x] = min(low[x], low[to]);
        }
        else if(!sccno[to])
            low[x] = min(low[x], dfn[to]);
    }
    if(dfn[x] == low[x]) {
        ++cnt1;
        while(true) {
            int now = ss.top();
            ss.pop();
            sccno[now] = cnt1;
            sd[now] = x;
            if(now == x)
                break;
            val[x] += val[now];            
        }
    }
}
void topo() {
    for(int i = 1; i <= n; i++) {
        if(!in[i] && sd[i] == i) {
            q.push(i);
            dis[i] = val[i];
        }
    }
    while(!q.empty()) {
        int now = q.front();
        q.pop();
        for(int i = head2[now]; i; i = edge2[i].nextt) {
            int to = edge2[i].to;
            dis[to] = max(dis[to], dis[now] + val[to]);
            if(!--in[to])
                q.push(to);
        }
    }
    for(int i = 1; i <= n; i++)
        ans = max(ans, dis[i]);
}
int main() {
    cin >> n >> m;
    for(int i = 1; i <= n; i++)
        cin >> val[i];
    for(int i = 1; i <= m; i++) {
        cin >> u >> v;
        addedge1(u, v);
    }
    for(int i = 1; i <= n; i++) {
        if(!dfn[i])
            tarjan(i);
    }
    for(int i = 1; i <= m; i++) {
        int u = sd[edge1[i].u], v = sd[edge1[i].to];
        if(u == v)
            continue;
        addedge2(u, v);
        in[v] ++;
    }
    topo();
    cout << ans << endl;
    return 0;
}

Luogu 3627 抢掠计划

先缩点，后跑一遍$spfa$（最长路）即可。

#include <bits/stdc++.h>
const int MAXN = 500050;
const int INF = 1 << 30;
using namespace std;
struct Edge1 {
    int u, to, nextt;
}edge1[MAXN];
struct Edge2 {
    int u, to, val, nextt;
}edge2[MAXN];
stack<int> d;
queue<int> q;
int n, m, u, v, w, x, num1, num2, cnt, cnt1, s, p, ans, sd[MAXN], head1[MAXN], head2[MAXN], low[MAXN], dfn[MAXN], vall[MAXN], dis[MAXN], sccno[MAXN];
bool vis[MAXN];
int read() {
    int x = 0;
    bool sign = false;
    char alpha = 0;
    while(!isdigit(alpha)) {
        sign |= alpha == '-';
        alpha = getchar();
    }
    while(isdigit(alpha)) {
        x = (x << 1) + (x << 3) + (alpha ^ 48);
        alpha = getchar();
    }
    return sign ? -x : x;
}
void addedge1(int u, int v) {
    edge1[++num1].u = u;
    edge1[num1].to = v;
    edge1[num1].nextt = head1[u];
    head1[u] = num1;
}
void addedge2(int u, int v, int w) {
    edge2[++num2].u = u;
    edge2[num2].to = v;
    edge2[num2].val = w;
    edge2[num2].nextt = head2[u];
    head2[u] = num2;
}
void tarjan(int x) {
    dfn[x] = low[x] = ++cnt;
    d.push(x);
    for(int i = head1[x]; i; i = edge1[i].nextt) {
        int to = edge1[i].to;
        if(!dfn[to]) {
            tarjan(to);
            low[x] = min(low[x], low[to]);
        }
        else if(!sccno[to])
            low[x] = min(low[x], dfn[to]);
    }
    if(dfn[x] == low[x]) { 
        ++cnt1;
        while(true) {
            int now = d.top();
            d.pop();
            sccno[now] = cnt1;
            sd[now] = x;
            if(x == now)
                break;
            vall[x] += vall[now];
        }
    }
}
void spfa() {
    dis[sd[s]] = vall[sd[s]];
    q.push(sd[s]);
    while(!q.empty()) {
        int now = q.front();
        q.pop();
        vis[now] = false;
        for(int i = head2[now]; i; i = edge2[i].nextt) {
            int to = edge2[i].to, val = edge2[i].val;
            if(dis[to] < dis[now] + val) {
                dis[to] = dis[now] + val;
                if(!vis[to]) {
                    vis[to] = true;
                    q.push(to);
                }
            }
        }
    }    
}
int main() {
    n = read();
    m = read();
    for(int i = 1; i <= m; i++) {
        u = read();
        v = read();
        addedge1(u, v);
    }
    for(int i = 1; i <= n; i++)
        vall[i] = read();
    for(int i = 1; i <= n; i++) {
        if(!dfn[i])
            tarjan(i);
    }
    for(int i = 1; i <= m; i++) {
        u = sd[edge1[i].u];
        v = sd[edge1[i].to];
        w = vall[v];
        if(u == v)
            continue;
        addedge2(u, v, w);
    }
    s = read();
    p = read();
    spfa();
    for(int i = 1; i <= p; i++) {
        x = read();
        ans = max(ans, dis[sd[x]]);
    }
    cout << ans << endl;
    return 0;
}

Luogu 2002 消息扩散

答案显然是缩点后入度为$0$的点的个数。

#include <bits/stdc++.h>
const int MAXN = 100050;
const int MAXM = 500500;
using namespace std;
stack<int> ss;
int n, m, u, v, pos, num, num1, num2, cnt, cnt1, in[MAXN], head1[MAXN], head2[MAXN], low[MAXN], dfn[MAXN], sd[MAXN], sccno[MAXN];
bool vis[5050][5050];
struct Edge1 {
    int u, to, nextt;
}edge1[MAXM];
struct Edge2 {
    int u, to, nextt;
}edge2[MAXM];
int read() {
    int x = 0;
    bool sign = false;
    char alpha = 0;
    while(!isdigit(alpha)) {
        sign |= alpha == '-';
        alpha = getchar();
    }
    while(isdigit(alpha)) {
        x = (x << 1) + (x << 3) + (alpha ^ 48);
        alpha = getchar();
    }
    return sign ? -x : x;
}
void addedge1(int u, int v) {
    edge1[++num1].u = u;
    edge1[num1].to = v;
    edge1[num1].nextt = head1[u];
    head1[u] = num1;
}
void addedge2(int u, int v) {
    edge2[++num2].u = u;
    edge2[num2].to = v;
    edge2[num2].nextt = head2[u];
    head2[u] = num2;
}
void tarjan(int x) {
    dfn[x] = low[x] = ++cnt;
    ss.push(x);
    for(int i = head1[x]; i; i = edge1[i].nextt) {
        int to = edge1[i].to;
        if(!dfn[to]) {
            tarjan(to);
            low[x] = min(low[x], low[to]);
        }
        else if(!sccno[to])
            low[x] = min(low[x], dfn[to]);
    }
    if(dfn[x] == low[x]) {
        ++cnt1;
        while(true) {
            int now = ss.top();
            ss.pop();
            sccno[now] = cnt1;
            sd[now] = cnt1;
            if(now == x)
                break;        
        }
    }
}
int main() {
    n = read(); 
    m = read();
    for(int i = 1; i <= m; i++) {
        u = read();
        v = read();
        if(u == v)
            continue;
        addedge1(u, v);
    }
    for(int i = 1; i <= n; i++) {
        if(!dfn[i])
            tarjan(i);
    }
    for(int i = 1; i <= num1; i++) {
        u = sd[edge1[i].u];
        v = sd[edge1[i].to];
        if(u == v)
            continue;
        in[v]++;
        addedge2(u, v);
    }
    for(int i = 1; i <= cnt1; i++) {
        if(!in[i])
            num++;
    }
    cout << num << endl;
    return 0;
}

Luogu 2746 Network of Schools

第一问的答案显然是缩点后入度为$0$的点的个数。

对于第二问，若存在出度为$0$的点或者入度为$0$的点则不满足条件。若把一个出度为$0$的点与一个入度为$0$的点连一条边，问题便迎刃而解。

考虑到两种点的个数不一定相同，选取其中的最大值即可。

记得判断是一个大环的情况。

#include <bits/stdc++.h>
const int MAXN = 250000;
using namespace std;
struct Edge1 {
    int u, to, nextt;
}edge1[MAXN << 1];
struct Edge2 {
    int u, to, nextt;
}edge2[MAXN << 1];
stack<int> s;
int n, u ,v, cnt, cnt1, ans1, num, num1, num2, head1[MAXN], head2[MAXN], dfn[MAXN], low[MAXN], sccno[MAXN], sd[MAXN], in[MAXN], out[MAXN];
void addedge1(int u, int v) {
    edge1[++num1].u = u;
    edge1[num1].to = v;
    edge1[num1].nextt = head1[u];
    head1[u] = num1;
}
void addedge2(int u, int v) {
    edge2[++num2].u = u;
    edge2[num2].to = v;
    edge2[num2].nextt = head2[u];
    head2[u] = num2;
}
void tarjan(int x) {
    dfn[x] = low[x] = ++cnt;
    s.push(x);
    for(int i = head1[x]; i; i = edge1[i].nextt) {
        int to = edge1[i].to;
        if(!dfn[to]) {
            tarjan(to);
            low[x] = min(low[x], low[to]);
        }
        else if(!sccno[to])
            low[x] = min(low[x], dfn[to]);
    }
    if(low[x] == dfn[x]) {
        ++cnt1;
        while(true) {
            int now = s.top();
            s.pop();
            sd[now] = cnt1;
            sccno[now] = cnt1;
            if(now == x)
                break;
        }
    }
}
int main() {
    cin >> n;
    for(int i = 1; i <= n; i++) {
        while(scanf("%d", &v) == 1) {
            if(v == 0)
                break;
            addedge1(i, v);
        }
    }
    for(int i = 1; i <= n; i++) {
        if(!dfn[i])
            tarjan(i);
    }
    for(int i = 1; i <= num1; i++) {
        u = sd[edge1[i].u];
        v = sd[edge1[i].to];
        if(u == v)
            continue;
        addedge2(u, v);
        in[v]++;
        out[u]++;
    }
    for(int i = 1; i <= cnt1; i++) {
        if(in[i] == 0)
            ans1++;
        if(out[i] == 0)
            num++;
    }
    cout << ans1 << endl;
    if(cnt1 == 1) {
        cout << "0" << endl;
        return 0;
    }
    cout << max(num, ans1) << endl;
    return 0;
} 

Luogu 2341 受欢迎的牛

若出现个数 ge 1的出度为$0$的强连通分量，则无解。

$otherwise$答案即为出度为$0$的强连通分量的大小。

#include <bits/stdc++.h>
const int MAXN = 10050;
const int MAXM = 50050;
using namespace std;
stack<int> ss;
int n, m, u, v, pos, num, num1, num2, cnt, cnt1, head1[MAXN], head2[MAXN], low[MAXN], dfn[MAXN], sd[MAXN], f[MAXN], sccno[MAXN];
struct Edge1 {
    int u, to, nextt;
}edge1[MAXM];
struct Edge2 {
    int u, to, nextt;
}edge2[MAXM];
int read() {
    int x = 0;
    bool sign = false;
    char alpha = 0;
    while(!isdigit(alpha)) {
        sign |= alpha == '-';
        alpha = getchar();
    }
    while(isdigit(alpha)) {
        x = (x << 1) + (x << 3) + (alpha ^ 48);
        alpha = getchar();
    }
    return sign ? -x : x;
}
void addedge1(int u, int v) {
    edge1[++num1].u = u;
    edge1[num1].to = v;
    edge1[num1].nextt = head1[u];
    head1[u] = num1;
}
void addedge2(int u, int v) {
    edge2[++num2].u = u;
    edge2[num2].to = v;
    edge2[num2].nextt = head2[u];
    head2[u] = num2;
}
void tarjan(int x) {
    dfn[x] = low[x] = ++cnt;
    ss.push(x);
    for(int i = head1[x]; i; i = edge1[i].nextt) {
        int to = edge1[i].to;
        if(!dfn[to]) {
            tarjan(to);
            low[x] = min(low[x], low[to]);
        }
        else if(!sccno[to])
            low[x] = min(low[x], dfn[to]);
    }
    if(dfn[x] == low[x]) {
        ++cnt1;
        while(true) {
            int now = ss.top();
            ss.pop();
            sccno[now] = cnt1;
            sd[now] = cnt1;
            f[cnt1] ++;
            if(now == x)
                break;        
        }
    }
}
int main() {
    n = read(); 
    m = read();
    for(int i = 1; i <= m; i++) {
        u = read();
        v = read();
        addedge1(u, v);
    }
    for(int i = 1; i <= n; i++) {
        if(!dfn[i])
            tarjan(i);
    }
    for(int i = 1; i <= m; i++) {
        u = sd[edge1[i].u];
        v = sd[edge1[i].to];
        if(u == v)
            continue;
        addedge2(u, v);
    }
    for(int i = 1; i <= cnt1; i++) {
        if(!head2[i]) {
            pos = i;
            num++;
        }
        if(num == 2) {
            cout << "0" << endl;
            return 0;
        }
    }
    cout << f[pos] << endl;
    return 0;
}

Luogu 1262 间谍网络

考虑无解的情况。若存在一个点既不能被贿赂，也没有入度，则不存在答案。

先缩点。对于现在没有入度且原先在强连通分量中的点，代价即为这些点中花费最小的值，累加即可。

#include <bits/stdc++.h>
const int MAXN = 100050;
const int INF = 1 << 30;
int n, m, u, v, p, num1, num2, cnt, cnt1, ans, in[MAXN], dfn[MAXN], low[MAXN], head1[MAXN], head2[MAXN], sccno[MAXN], val[MAXN], vall[MAXN], sd[MAXN];
using namespace std;
stack<int> s;
struct Edge1 {
    int u, to, nextt;
}edge1[MAXN];
int read() {
    int x = 0;
    bool sign = false;
    char alpha = 0;
    while(!isdigit(alpha)) {
        sign |= alpha == '-';
        alpha = getchar();
    }
    while(isdigit(alpha)) {
        x = (x << 1) + (x << 3) + (alpha ^ 48);
        alpha = getchar();
    }
    return sign ? -x : x;
}
void addedge1(int u, int v) {
    edge1[++num1].u = u;
    edge1[num1].to = v;
    edge1[num1].nextt = head1[u];
    head1[u] = num1;
}
void tarjan(int x) {
    dfn[x] = low[x] = ++cnt;
    s.push(x);
    for(int i = head1[x]; i; i = edge1[i].nextt) {
        int to = edge1[i].to;
        if(!dfn[to]) {
            tarjan(to);
            low[x] = min(low[x], low[to]);
        }
        else if(!sccno[to])
            low[x] = min(low[x], dfn[to]);
    }
    if(dfn[x] == low[x]) {
        ++cnt1;
        while(true) {
            int now = s.top();
            s.pop();
            sd[now] = cnt1;
            sccno[now] = cnt1;
            vall[cnt1] = min(vall[cnt1], val[now]);
            if(now == x)
                break;
        }
    }
}
int main() {
    n = read();
    p = read();
    for(int i = 1; i <= n; i++)
        val[i] = vall[i] = INF;
    for(int i = 1; i <= p; i++) {
        u = read();
        val[u] = read();
    }
    m = read();
    for(int i = 1; i <= m; i++) {
        u = read();
        v = read();
        in[v]++;
        addedge1(u, v);
    }
    for(int i = 1; i <= n; i++) {
        if(val[i] == INF && in[i] == 0) {
            cout << "NO" << endl << i << endl;
            return 0;
        }
    }
    for(int i = 1; i <= n; i++) {
        if(!dfn[i] && val[i] != INF)
            tarjan(i);
    }
    for(int i = 1; i <= n; i++)
        in[i] = 0;
    for(int i = 1; i <= num1; i++) {
        u = sd[edge1[i].u];
        v = sd[edge1[i].to];
        if(u == v)
            continue;
        in[v]++;
    }
    for(int i = 1; i <= cnt1; i++) {
        if(in[i] == 0)
            ans += vall[i];
    }
    cout << "YES" << endl << ans << endl;
    return 0;
}