桥与割点 - tarjan
桥判定法则:无向边(x,y)是桥,当且仅当搜索树上 存在 x 的一个子节点y,满足
\[dfn[x] <low[y]
\]
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <stack>
using namespace std;
#define debug(x) cerr << #x << "=" << x << endl;
const int MAXN = 100000 + 10;
const int MAXM = 100000 * 2 + 10;
int n,m,last[MAXN],edge_tot=1,root,tarjan_cnt,dfn[MAXN],low[MAXN],ans;
bool flg[MAXN];
struct Edge{
int u,v,to;
Edge(){}
Edge(int u, int v, int to) : u(u), v(v), to(to) {}
}e[MAXM];
inline void add(int u, int v) {
e[++edge_tot] = Edge(u, v, last[u]);
last[u] = edge_tot;
}
void tarjan(int x, int in_edge) {
dfn[x] = low[x] = ++tarjan_cnt;
int num = 0;
for(int i=last[x]; i; i=e[i].to) {
int v = e[i].v;
if(!dfn[v]) {
tarjan(v, i);
low[x] = min(low[x], low[v]);
if(low[v] > dfn[x]) {
bridge[i] = bridge[i^1] = true;
}
} else if(i != in_edge ^ 1){
low[x] = min(low[x], dfn[v]);
}
}
}
int main() {
scanf("%d%d", &n, &m);
for(int i=1; i<=m; i++) {
int xi, yi;
scanf("%d%d", &xi, &yi);
add(xi, yi);
add(yi, xi);
}
for(int i=1; i<=n; i++) {
if(!dfn[i]) {
tarjan(i, 0);
}
}
return 0;
}
割点判定法则:若x不为搜索树的根,当且仅当搜索树上存在x的一个子节点y,满足
\[dfn[x] \leq low[y]
\]
特别地,若x为搜索树的根,当且仅当搜索树上存在至少两个x的子节点满足上述条件
由于是小于等于号,所以不用管重边与遍历到父节点什么的
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <stack>
using namespace std;
#define debug(x) cerr << #x << "=" << x << endl;
const int MAXN = 100000 + 10;
const int MAXM = 100000 * 2 + 10;
int n,m,last[MAXN],edge_tot,root,tarjan_cnt,dfn[MAXN],low[MAXN],ans;
bool flg[MAXN];
struct Edge{
int u,v,to;
Edge(){}
Edge(int u, int v, int to) : u(u), v(v), to(to) {}
}e[MAXM];
inline void add(int u, int v) {
e[++edge_tot] = Edge(u, v, last[u]);
last[u] = edge_tot;
}
void tarjan(int x) {
dfn[x] = low[x] = ++tarjan_cnt;
int num = 0;
for(int i=last[x]; i; i=e[i].to) {
int v = e[i].v;
if(!dfn[v]) {
tarjan(v);
low[x] = min(low[x], low[v]);
if(low[v] >= dfn[x]) {
num++;
if(x != root || num > 1) flg[x] = true; // 不要在这里计数, 若非要在这里计数,还要加上一个 !flg[x]
}
} else {
low[x] = min(low[x], dfn[v]);
}
}
}
int main() {
scanf("%d%d", &n, &m);
for(int i=1; i<=m; i++) {
int xi, yi;
scanf("%d%d", &xi, &yi);
add(xi, yi);
add(yi, xi);
}
for(int i=1; i<=n; i++) {
if(!dfn[i]) {
root = i;
tarjan(i);
}
}
for(int i=1; i<=n; i++) { //在这里计数最好
if(flg[i]) ans++;
}
printf("%d\n", ans);
for(int i=1; i<=n; i++) {
if(flg[i]) printf("%d ", i);
}
return 0;
}