洛谷 [P3388] 割点模版

tarjan 求无向图的割点

割点,即割去此点后原图可变为两个或多个独立的联通块
一个点 x 是割点,当且仅当存在一个x 的子节点 y ,使得 low[y] >= dfn[x]
对于根节点来说,需要两个满足的节点

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int MAXN = 100005;
int head[MAXN], nume, rot, n, m, dfn[MAXN], low[MAXN], ind, cnt;
bool f[MAXN];
struct edge{
	int to, nxt;
}e[MAXN << 1];
void adde(int from, int to) {
	e[++nume].to = to;
	e[nume].nxt = head[from];
	head[from] = nume;
}
int init() {
	int rv = 0, fh = 1;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') fh = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		rv = (rv<<1) + (rv<<3) + c - '0';
		c = getchar();
	}
	return fh * rv;
}
void tarjan(int u) {
	dfn[u] = low[u] = ++ind;
	int flag = 0;
	for(int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if(!dfn[v]) {
			tarjan(v);
			low[u] = min(low[u], low[v]);
			if(low[v] >= dfn[u]) {
				flag++;
				if(u != rot || flag > 1) f[u] = 1;
			}
		}else low[u] = min(low[u], dfn[v]);
	}
}
int main() {
	n = init(); m = init();
	for(int i = 1; i <= m; i++) {
		int u = init(), v = init();
		adde(u, v); adde(v, u);
	}
	for(int i = 1; i <= n; i++) if(!dfn[i]) rot = i, tarjan(i);
	for(int i = 1; i <= n; i++) if(f[i]) cnt++;
	cout << cnt << endl;
	for(int i = 1; i <= n; i++) if(f[i]) printf("%d ", i);
	printf("\n");
	return 0;
}
posted @ 2018-04-02 10:44  Mr_Wolfram  阅读(120)  评论(0编辑  收藏