【tarjan强连通分量】洛谷P1726 上白泽慧音

【tarjan强连通分量】洛谷P1726 上白泽慧音

题目传送门

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妥妥的强连通模板啊(详细解释戳这里)

#include <bits/stdc++.h>
#define MAXN 5005
#define MAXM 50005*2
using namespace std;

int n, m, cnt;
int head[MAXN], Next[MAXM], vet[MAXM];
vector<int> ans;
//链式前向星存边
void add(int x, int y){
    cnt++;
    Next[cnt] = head[x];
    vet[cnt] = y;
    head[x] = cnt;
}

int T, dfn[MAXN], low[MAXN];          //时间戳,子树中连通的最小dfn
bool vis[MAXN];             //判断点是否在栈内
stack<int> s;
void tarjan(int x){
    T++;
    dfn[x] = low[x] = T;
    s.push(x);
    vis[x] = true;
    for (int i = head[x]; i; i = Next[i]) {
        int v = vet[i];
        if(!dfn[v]){            //没有访问过,为搜索树的一条边,需要继续搜索
            tarjan(v);
            low[x] = min(low[x], low[v]);       //回溯赋值
        }
        else if(vis[v]){           //v在栈中,即为返祖边或者横叉边
            low[x] = min(low[x], dfn[v]);
        }
    }
    if(dfn[x] == low[x]){           //是一个强连通分量的搜索树根
        int t = -1;
        vector<int> vec;
        vec.clear();
        while(t != x){              //出栈直到x
            t = s.top();
            vec.push_back(t);
            s.pop();
            vis[t] = false;
        }
        if(vec.size() > ans.size()){        //记录答案
            ans = vec;
        }
    }
}

int main()
{
    cin >> n >> m;
    int a,b,t;
    for (int i = 1; i <= m; ++i) {
        scanf("%d %d %d", &a, &b, &t);
        if(a == b)
            continue;
        add(a,b);
        if(t == 2){
            add(b,a);
        }
    }
    for (int i = 1; i <= n; ++i) {      //搜索强连通分量
        if(!dfn[i])
            tarjan(i);
    }
    cout << ans.size() << endl;
    sort(ans.begin(), ans.end());
    for (int i = 0; i < ans.size(); ++i) {
        cout << ans[i] << " ";
    }
    cout << endl;

    return 0;
}