tarjan算法板子

無向圖

• 概念

  • 時間戳
    \(dfn[x]\),在深度優先遍歷中，按照每個節點第一次被訪問的順序，依次做整數標記

  • 追溯值
    \(low[x]\),通過非搜索邊能到達的最小時間戳

割邊判定法則

• 無向邊\((x,y)\)是割邊/橋，當且僅當存在x的一個子節點滿足\(dfn[x] < low[y]\)
  刪除無向邊\((x,y)\)后，圖斷開成兩個部分

Code

int dfn[N], low[N], dfcnt;
bool g[M];
void tarjan(int x, int ei) {
    dfn[x] = low[x] = ++dfcnt;
    for(int i = head[x]; i; i = e[i].next) {
        int y = e[i].t;
        if (!dfn[y]) {
            tarjan(y, i);
            low[x] = min(low[x], low[y]);
            if (dfn[x] < low[y]) g[i] = g[i^1] = 1;
        }
        else if (i != (ei^1)) low[x] = min(low[x], dfn[y]);
    }
}

割點判定法則

• 若x不是根節點，則x是割點當且僅當存在一個子節點y滿足\(dfn[x]\leq low[y]\)
  若x是根節點，則x是割點當且僅當存在至少兩個子節點\(y_1,y_2\)滿足上條件

Code

int dfn[N], low[N], dfcnt, rt;
bool g[N];
void tarjan(int x) {
    dfn[x] = low[x] = ++dfcnt;
    int son = 0;
    for (int i = head[x]; i; i = e[i].next) {
        int y = e[i].t;
        if (!dfn[y]) {
            tarjan(y);
            low[x] = min(low[x], low[y]);
            if (dfn[x] <= low[y]) {
                son++;
                if (x != rt || son > 1) g[x] = 1;
            }
        }
        else low[x] = min(low[x], dfn[y]);
    }
}

點雙聯通分量

• 對於，每個點雙中來說，圖里是不存在割點的
  於是，這里可以就可以將圖轉成一顆圓方樹了。

Code

int dfn[N], low[N], dfcnt, sta[N], top, cnt;
vector<int> dcc[N];
bool  g[N];
void tarjan(int x, int rt) {
    dfn[x] = low[x] = ++dfcnt;
    sta[++top] = x;
    int son = 0;
    for (int i = head[x]; i; i = e[i].next) {
        int y = e[i].t;
        if (!dfn[y]) {
            tarjan(y); 
            low[x] = min(low[x], low[y]);
            if (dfn[x] <= low[y]) {
                son++;
                if (x != rt || son > 1)  g[x] = 1;
                dcc[++cnt].clear();
                while (1) {
                    int z = sta[top--];
                    dcc[cnt].push_back(z);
                    if (y == z) break;
                }
                dcc[cnt].push_back(x);
            }
        }
        else low[x] = min(low[x], dfn[y]);
    }
}

邊雙聯通分量

• 對於一個邊雙，任意兩個點都有兩條不重合的路徑

Code

int dfn[N], low[N], dfcnt;
bool g[M];
void tarjan(int x, int ei) {
    dfn[x] = low[x] = ++dfcnt;
    for(int i = head[x]; i; i = e[i].next) {
        int y = e[i].t;
        if (!dfn[y]) {
            tarjan(y, i);
            low[x] = min(low[x], low[y]);
            if (dfn[x] < low[y]) g[i] = g[i^1] = 1;
        }
        else if (i != (ei^1)) low[x] = min(low[x], dfn[y]);
    }
}
int n, m, d[N], b[N], cnt, ans;
void dfs(int x) {
    b[x] = cnt;
    for(int i = head[x]; i; i = e[i].next) {
        int y = e[i].t;
        if (b[y] || g[i]) continue;
        dfs(y);
    }
}
int main() {
    //~~~
    for(int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i, 0);
    for(int i = 1; i <= n; i++)
        if (!b[i]) cnt++, dfs(i);
    //~~~
    return 0;
}

有向圖

有向圖的強聯通分量

• 在一個強聯通分量中，存在x到y的路徑，就存在y到x的路徑

Code

void tarjan(int x) {
    dfn[x] = low[x] = ++dfcnt;
    s[++top] = x;
    for(int i = head[x]; i; i = e[i].next) {
        int y = e[i].t;
        if (!dfn[y]) tarjan(y), low[x] = min(low[x], low[y]);
        else if (!b[y]) low[x] = min(low[x], dfn[y]);
    }
    if (dfn[x] == low[x]) {
        cnt++;
        while(1) {
            int y = s[top--];
            b[y] = cnt;
            size[cnt]++;
            if (x == y) break;
        }
    }
}

例題

• luoguP3387縮點

Code

#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1e4+5, M = 1e5+5;
struct side { int t, next; } e[M][2];
int head[N][2], tot[2];
void add(int x, int y, int k) {
    e[++tot[k]][k].next = head[x][k]; 
    head[x][k] = tot[k];
    e[tot[k]][k].t = y;
}
int n, m, w[N], r[N], d[N], ans;
int dfn[N], low[N], dfcnt, sta[N], top, cnt, bel[N], sum[N];
void tarjan(int x) {
    dfn[x] = low[x] = ++dfcnt;
    sta[++top] = x;
    for (int i = head[x][0]; i; i = e[i][0].next) {
        int y = e[i][0].t;
        if (!dfn[y]) tarjan(y), low[x] = min(low[x], low[y]);
        else if (!bel[y]) low[x] = min(low[x], dfn[y]);
    }
    if (dfn[x] == low[x]) {
        cnt++;
        while (1) {
            int y = sta[top--];
            bel[y] = cnt;
            sum[cnt] += w[y];
            if (x == y) break;
        }
    }
}
queue<int> q;
int tuopu() {
    for (int i = 1; i <= cnt; i++)
        if (!r[i]) q.push(i), d[i] = sum[i];
    while (!q.empty()) {
        int x = q.front(); q.pop();
        for (int i = head[x][1]; i; i = e[i][1].next) {
            int y = e[i][1].t;
            d[y] = max(d[y], d[x] + sum[y]);
            if (--r[y] == 0) q.push(y);
        }
    }
    for (int i = 1; i <= cnt; i++)
        ans = max(ans, d[i]);
    return ans;
}
int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%d", &w[i]);
    for (int i = 1; i <= m; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y, 0);
    }
    for (int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i);
    for (int x = 1; x <= n; x++)
        for (int i = head[x][0]; i; i = e[i][0].next) {
            int y = e[i][0].t;
            if (bel[x] != bel[y]) 
                r[bel[y]]++, add(bel[x], bel[y], 1);
        }
    printf("%d\n", tuopu());
    return 0;
}