学渣乱搞系列之Tarjan模板合集
by 狂徒归来
一、求强连通子图
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #include <stack> 13 #define LL long long 14 #define pii pair<int,int> 15 #define INF 0x3f3f3f3f 16 using namespace std; 17 const int maxn = 20100; 18 int dfn[maxn],low[maxn],belong[maxn]; 19 bool instack[maxn]; 20 vector<int>g[maxn]; 21 stack<int>stk; 22 int n,m,cnt,scc; 23 void tarjan(int u) { 24 dfn[u] = low[u] = ++cnt; 25 stk.push(u); 26 instack[u] = true; 27 for(int i = 0; i < g[u].size(); i++) { 28 if(!dfn[g[u][i]]) { 29 tarjan(g[u][i]); 30 low[u] = min(low[u],low[g[u][i]]); 31 } else if(instack[g[u][i]]) low[u] = min(low[u],dfn[g[u][i]]); 32 } 33 if(dfn[u] == low[u]) { 34 scc++; 35 int v; 36 do { 37 v = stk.top(); 38 instack[v] = false; 39 belong[v] = scc; 40 stk.pop(); 41 } while(v != u);//每一个强连通块内de点，以及其属于的scc 42 } 43 } 44 int main() { 45 int i,j,u,v,a,b; 46 while(~scanf("%d %d",&n,&m)) { 47 for(i = 0; i <= n; i++) { 48 dfn[i] = belong[i] = 0; 49 instack[i] = false; 50 g[i].clear(); 51 } 52 cnt = scc = 0; 53 while(!stk.empty()) stk.pop(); 54 for(i = 1; i <= m; i++) { 55 scanf("%d %d",&u,&v); 56 g[u].push_back(v); 57 } 58 for(i = 1; i <= n; i++) 59 if(!dfn[i]) tarjan(i); 60 } 61 return 0; 62 }
二、求割点
1 const int maxn = 1010; 2 vector<int>g[maxn]; 3 bool iscut[maxn]; 4 int dfn[maxn],low[maxn],cnt,vis[maxn]; 5 void tarjan(int u,int fa) { 6 dfn[u] = low[u] = ++cnt; 7 vis[u] = 1; 8 int son = 0; 9 for(int i = 0; i < g[u].size(); i++) { 10 if(!vis[g[u][i]]) { 11 tarjan(g[u][i],u); 12 son++; 13 low[u] = min(low[u],low[g[u][i]]); 14 if(fa == -1 && son > 1 || fa != -1 && low[g[u][i]] >= dfn[u]) 15 iscut[u] = true; 16 } else if(vis[g[u][i]] == 1) low[u] = min(low[u],dfn[g[u][i]]); 17 } 18 vis[u] = 2; 19 }
三、求割边/桥
1 int ret; 2 void tarjan(int u,int fa) { 3 dfn[u] = low[u] = ++clk; 4 bool flag = false; 5 for(int i = head[u]; ~i; i = e[i].next) { 6 if(!flag && e[i].to == fa) { 7 flag = true; 8 continue; 9 } 10 if(!dfn[e[i].to]) { 11 tarjan(e[i].to,u); 12 low[u] = min(low[u],low[e[i].to]); 13 if(low[e[i].to] > dfn[u]) { 14 e[i].cut = e[i^1].cut = true; 15 ++ret; 16 } 17 } else low[u] = min(low[u],dfn[e[i].to]); 18 } 19 }
四、求LCA