It’s the year 5555. You have a graph, and you want to find a long cycle and a huge independent set, just because you can. But for now, let’s just stick with finding either.

Given a connected graph with n vertices, you can choose to either:

find an independent set that has exactly ?n??√? vertices.
find a simple cycle of length at least ?n??√?.
An independent set is a set of vertices such that no two of them are connected by an edge. A simple cycle is a cycle that doesn’t contain any vertex twice. I have a proof you can always solve one of these problems, but it’s too long to fit this margin.

Input
The first line contains two integers n and m (5≤n≤105, n?1≤m≤2?105) — the number of vertices and edges in the graph.

Each of the next m lines contains two space-separated integers u and v (1≤u,v≤n) that mean there’s an edge between vertices u and v. It’s guaranteed that the graph is connected and doesn’t contain any self-loops or multiple edges.

Output
If you choose to solve the first problem, then on the first line print “1”, followed by a line containing ?n??√? distinct integers not exceeding n, the vertices in the desired independent set.

If you, however, choose to solve the second problem, then on the first line print “2”, followed by a line containing one integer, c, representing the length of the found cycle, followed by a line containing c distinct integers integers not exceeding n, the vertices in the desired cycle, in the order they appear in the cycle.

Examples
inputCopy
6 6
1 3
3 4
4 2
2 6
5 6
5 1
outputCopy
1
1 6 4
inputCopy
6 8
1 3
3 4
4 2
2 6
5 6
5 1
1 4
2 5
outputCopy
2
4
1 5 2 4
inputCopy
5 4
1 2
1 3
2 4
2 5
outputCopy
1
3 4 5

思路：和codeforces 649div2 D題差不多，都是dfs樹的思路，但是這個題目比較難一些。
這個題目是dfs樹找最大環。如果最大環符合第二個條件的話，就輸出這個樹中的點。如果不符合的話，就需要討論了。
圖片來源

代碼如下：

#include<bits/stdc++.h> #define ll long long #define inf 0x3f3f3f3f using namespace std;const int maxx=1e5+100; const int maxm=2e5+100; struct edge{int to,next; }e[maxm<<1]; int head[maxm<<1]; int f[maxx],dis[maxx],vis[maxx]; vector<int> p[maxx]; int n,m,tot,ans=0,pos;inline void init() {memset(f,0,sizeof(f));memset(dis,0,sizeof(dis));memset(vis,0,sizeof(vis));memset(head,-1,sizeof(head));tot=0; } inline void add(int u,int v) {e[tot].to=v,e[tot].next=head[u],head[u]=tot++; } inline void dfs(int u,int fa,int h,int cor,int num) {vis[u]=1;f[u]=fa;dis[u]=h;p[h%(num-1)].push_back(u);for(int i=head[u];i!=-1;i=e[i].next){int to=e[i].to;if(to==fa) continue;else if(vis[to]){if(ans<abs(dis[u]-dis[to])+1){ans=abs(dis[u]-dis[to])+1;pos=u;}}else dfs(to,u,h+1,cor^1,num);} } int main() {scanf("%d%d",&n,&m);init();int x,y;for(int i=1;i<=m;i++){scanf("%d%d",&x,&y);add(x,y);add(y,x);}int num=sqrt(n);if(num*num<n) num+=1;dfs(1,0,0,0,num);if(m==n-1||ans<num){cout<<1<<endl;for(int i=0;i<num;i++){if(p[i].size()>=num){for(int j=0;j<num;j++) cout<<p[i][j]<<" ";cout<<endl;return 0;}}}else{cout<<2<<endl<<ans<<endl;for(;ans;ans--) cout<<pos<<" ",pos=f[pos];cout<<endl;} }

努力加油a啊，(^o)/~

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