思路
dfs 树的一道出色的应用题
令 \(k=\lceil \sqrt n \rceil\)
我们先按照遍历的顺序构建出 dfs 树
对于一条返祖边 \((u, v)\),如果有 \(dep_u-dep_v +1\ge k\),那么 dfs 树上的链 \((v, u)\) 就是一个满足要求的环
假如并没有满足要求的环,说明对于从根出发的链上任意两点 \((x,y)\),如果有 \(dep_x+k-1\ge dep_y\),那么在原图中 \(x,y\) 一定没有连边
而且 dfs 树上不位于同一条链上的两点之间也没有连边
于是我们可以按 \(dep\bmod (k-1)\) 进行分组
根据抽屉原理,一定有一个组点的个数 \(\ge k\)
#include<iostream>
#include<fstream>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#define LL long long
#define FOR(i, x, y) for(int i = (x); i <= (y); i++)
#define ROF(i, x, y) for(int i = (x); i >= (y); i--)
#define PFOR(i, x) for(int i = he[x]; i; i = r[i].nxt)
inline int reads()
{
int sign = 1, re = 0; char c = getchar();
while(c < '0' || c > '9'){if(c == '-') sign = -1; c = getchar();}
while('0' <= c && c <= '9'){re = re * 10 + (c - '0'); c = getchar();}
return sign * re;
}
int n, m, k;
struct Node
{
int to, nxt;
}r[400005]; int he[100005];
std::bitset<200005> rvis;
inline void Edge_add(int u, int v)
{
static int cnt = 1;
r[++cnt] = (Node){v, he[u]};
he[u] = cnt;
}
std::bitset<100005> vis;
int dep[100005], fa[100005];
std::vector<int> q[100005];
void dfs(int now)
{
vis[now] = 1; q[dep[now] % (k - 1)].emplace_back(now);
PFOR(i, now)
{
if(rvis[i >> 1]) continue;
int to = r[i].to;
if(vis[to])
{
if(dep[now] - dep[to] + 1 >= k)
{
printf("2\n%d\n", dep[now] - dep[to] + 1);
while(1)
{
printf("%d ", now);
if(now == to) exit(0);
now = fa[now];
}
}
}
else
{
dep[to] = dep[now] + 1;
fa[to] = now;
rvis[i >> 1] = 1;
dfs(to);
}
}
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("test.in", "r", stdin);
freopen("test.out", "w", stdout);
#endif
n = reads(), m = reads(); k = ceil(sqrt(n));
FOR(i, 1, m)
{
int u = reads(), v = reads();
Edge_add(u, v), Edge_add(v, u);
}
FOR(i, 1, n) if(!vis[i])
dep[i] = 1,
dfs(i);
puts("1");
FOR(i, 0, m - 1) if(q[i].size() >= k)
{
FOR(j, 0, k - 1) printf("%d ", q[i][j]);
break;
}
return 0;
}
标签:CF1325F,include,Last,dep,dfs,int,reads,Ehab,now
From: https://www.cnblogs.com/zuytong/p/16661983.html