标签:clu queue ons http line phi -- time ace
这是一道用tarjan做的模板,要求找到有向图中最大的联通块。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl ‘\n‘ #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); //priority_queue<int ,vector<int>, greater<int> >que; const ll oo = 1ll<<17; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1000000007; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar(); while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar(); return x=f?-x:x; } /*-----------------------showtime----------------------*/ const int maxn = 1e5+9; struct E{ int u,v; int nxt; }edge[maxn]; int head[maxn],gtot = 0; void addedge(int u,int v){ edge[gtot].u = u; edge[gtot].v = v; edge[gtot].nxt = head[u]; head[u] = gtot++; } priority_queue<int,vector<int>,greater<int> >tmp,ans;//这是一个小根堆q int vis[maxn],dfn[maxn],low[maxn]; stack<int>sk; int tot = 0; void tarjan(int u){ tot++; dfn[u] = low[u] = tot; vis[u] = 1; sk.push(u); for(int i=head[u]; ~i; i = edge[i].nxt){ int v = edge[i].v; if(dfn[v] == 0) { tarjan(v); low[u] = min(low[u], low[v]); } else if(vis[v]){ low[u] = min(low[u], dfn[v]); } } if(dfn[u] == low[u]){ while(!tmp.empty())tmp.pop(); while(!sk.empty() && sk.top() != u){ vis[sk.top()] = 0; tmp.push(sk.top()); sk.pop(); } vis[sk.top()] = 0; tmp.push(sk.top()); sk.pop(); if(tmp.size() > ans.size()) ans = tmp; else if(tmp.size() == ans.size() && tmp.top() < ans.top()) ans = tmp; } } int main(){ int n,m; scanf("%d%d", &n, &m); memset(head, -1, sizeof(head)); for(int i=1; i<=m; i++){ int u,v,op; scanf("%d%d%d", &u, &v, &op); if(op == 1) addedge(u,v); else { addedge(u,v); addedge(v,u); } } for(int i=1; i<=n; i++) if(dfn[i] == 0)tarjan(i); printf("%d\n", (int)ans.size()); while(!ans.empty()){ printf("%d ", ans.top()); ans.pop(); } puts(""); return 0; }
标签:clu queue ons http line phi -- time ace
原文地址:https://www.cnblogs.com/ckxkexing/p/10351664.html