【模板】割点(割顶) Tarjan
题目背景
割点
题目描述
给出一个nnn个点,mmm条边的无向图,求图的割点。
输入输出格式
输入格式:第一行输入n,mn,mn,m
下面mmm行每行输入x,yx,yx,y表示xxx到yyy有一条边
输出格式:第一行输出割点个数
第二行按照节点编号从小到大输出节点,用空格隔开
输入输出样例
说明
对于全部数据,n≤20000n \le 20000n≤20000,m≤100000m \le 100000m≤100000
点的编号均大于000小于等于nnn。
tarjan图不一定联通。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<functional> #include<sstream> //#include<cctype> //#pragma GCC optimize(2) using namespace std; #define maxn 200005 #define inf 0x3f3f3f3f //#define INF 1e18 #define rdint(x) scanf("%d",&x) #define rdllt(x) scanf("%lld",&x) #define rdult(x) scanf("%lu",&x) #define rdlf(x) scanf("%lf",&x) #define rdstr(x) scanf("%s",x) typedef long long ll; typedef unsigned long long ull; typedef unsigned int U; #define ms(x) memset((x),0,sizeof(x)) const long long int mod = 1e9 + 7; #define Mod 1000000000 #define sq(x) (x)*(x) #define eps 1e-3 typedef pair<int, int> pii; #define pi acos(-1.0) const int N = 1005; #define REP(i,n) for(int i=0;i<(n);i++) typedef pair<int, int> pii; inline ll rd() { ll x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x; } ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); } ll sqr(ll x) { return x * x; } /*ll ans; ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans; } */ struct node { int to,nxt; }edge[maxn<<1]; int n, m; int idx, cnt, tot; int head[maxn], dnf[maxn], low[maxn]; bool cut[maxn]; void addedge(int u, int v) { edge[++cnt].to = v; edge[cnt].nxt= head[u]; head[u] = cnt; } void tarjan(int u, int fa) { dnf[u] = low[u] = ++idx; int ch = 0; for (int i = head[u]; i; i = edge[i].nxt) { int to = edge[i].to; if (!dnf[to]) { tarjan(to, fa); low[u] = min(low[u], low[to]); if (low[to] >= dnf[u] && u != fa) { cut[u] = true; } if (u == fa)ch++; } low[u] = min(low[u], dnf[to]); } if (ch >= 2 && u == fa)cut[u] = true; } int main() { //ios::sync_with_stdio(0); rdint(n); rdint(m); for (int i = 0; i < m; i++) { int a, b; rdint(a); rdint(b); addedge(a, b); addedge(b, a); } for (int i = 1; i <= n; i++) { if (dnf[i] == 0)tarjan(i, i); } for (int i = 1; i <= n; i++) { if (cut[i])tot++; } cout << tot << endl; for (int i = 1; i <= n; i++) { if (cut[i])cout << i << ' '; } return 0; }
EPFL - Fighting