图论 ---- Codeforces Round #649 (Div. 2)D题[dfs求环+深度分层求图中独立集]

D. Ehab’s Last Corollary

题目大意：

就是给你一个联通图，你有两种选择
1.你可以输出包含⌈k2⌉\lceil{k\over2}\rceil⌈2k⌉个顶点得独立点集，什么是独立点集：集合中任意两点都没有边相连
2.你要找到顶点数<=k<=k<=k的环

解题思路：

求环很简单就是dfs一遍如果深度大的点指向了深度小的点那么就是有环，现在是怎么求独立集？
我们一个想法就是将有边相连的点就将它们分开那么我们就可以按深度进形分层，如果两个点有边按照dfs序肯定不在同一层

#include <iostream>
#include <cstdio>
#include <stack>
#include <sstream>
#include <vector>
#include <map>
#include <cstring>
#include <deque>
#include <cmath>
#include <iomanip>
#include <queue>
#include <algorithm>
#include <set>
#define mid ((l + r) >> 1)
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define log2(a) log(a)/log(2)
#define _for(i,a,b) for( int i = (a); i < (b); ++i)
#define _rep(i,a,b) for( int i = (a); i <= (b); ++i)
#define for_(i,a,b) for( int i = (a); i >= (b); -- i)
#define rep_(i,a,b) for( int i = (a); i > (b); -- i)
#define lowbit(x) ((-x) & x)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x3f3f3f3f
#define hash Hash
#define next Next
#define count Count
#define pb push_back
#define f first
#define s second
using namespace std;
const int N = 4e5+10, mod = 1e9 + 7;
const double eps = 1e-10;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
template<typename T> void read(T &x)
{x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args)
{read(first);read(args...);
}
vector<int> G[N], ans, d[N];
int n, m, k;
int dep[N], path[N];void find(int u, int v)
{while(u != v){ans.pb(u);u = path[u];}ans.pb(u);cout << 2 << "\n";cout << ans.size() << "\n";for(auto it : ans)cout << it << " ";exit(0);return;
}void dfs(int deep, int u)
{dep[u] = deep;d[deep].pb(u);for(auto it : G[u]){if(dep[it] == -1) {path[it] = u;dfs(deep+1,it);}else if(it != path[u] && dep[u] - dep[it] + 1 <= k && dep[u] > dep[it]) find(u,it);}return;
}int main()
{ms(dep,-1);read(n,m,k);while(m --){int l, r;read(l,r);G[l].pb(r); G[r].pb(l);}dfs(0,1);int need = (k+1)/2;for(int i = 0; i <= n && need; i += 2)for(auto it : d[i]){ans.pb(it);if(--need == 0) break;}if(need){need = (k+1)/2;ans.clear();for(int i = 1; i <= n && need; i += 2)for(auto it : d[i]){ans.pb(it);if(--need == 0) break;}}cout << "1\n";for(auto i : ans)cout << i << " ";return 0;
}

图论 ---- Codeforces Round #649 (Div. 2)D题[dfs求环+深度分层求图中独立集]相关推荐

Codeforces Round 700 (Div. 2) B题英雄杀怪兽
Codeforces Round 700 (Div. 2) B题链接: https://codeforces.com/contest/1480/problem/B 大致意思: n组数据,每组数据的第 ...
Codeforces Round #774 (Div. 2)E题题解
Codeforces Round #774 (Div. 2) E. Power Board 题目陈述有一个n×m(1≤n,m≤106)n\times m(1\le n,m\le10^6)n×m(1≤ ...
Codeforces Round #649 (Div.2)题解
文章目录 A - XXXXX B - Most socially-distanced subsequence C - Ehab and Prefix MEXs A - XXXXX 题意:这个题让你找从 ...
Codeforces Round #807 (Div. 2)补题
C. Mark and His Unfinished Essay https://codeforces.com/contest/1705/problem/C 会卡long long,下面解法62ms过 ...
Codeforces Round #723 (Div. 2)补题
水题,只需要将序列分成两部分即可,一部分是大的,一部分是小的. #include <cstdio> #include <iostream> #include <algor ...
Codeforces Round #552 (Div. 3) C题
题目链接 http://codeforces.com/contest/1154/problem/C 题目 C. Gourmet Cat time limit per test 1 second mem ...
Codeforces Round #649 (Div. 2)C. Ehab and Prefix MEXs[排列的构造]
C. Ehab and Prefix MEXs 题目大意: 解题思路:题目说保证a数组是非递减的,那么如果某位置a[i]!=a[i−1]a[i]!=a[i-1]a[i]!=a[i−1]那么这个位置ii ...
双指针--Codeforces Round #645 (Div. 2) d题
D. The Best Vacation 题目大意: 算出连续x天最多的拥抱,一个月第i号就有i个拥抱思路:双指针,扫描过去(每个月每个月的计算,最后超出的部分再一天一天算) 代码 : #inclu ...
思维--找规律--Codeforces Round #645 (Div. 2) c题
C. Celex Update 题目大意:给出两点的坐标,找出不同的路径的总数(路径数字总和不同) 思路:根据观察向下走比向右走的增加幅度加1,所以在第i步向下对sum的影响是 n-i+1 所以最 ...

图论 ---- Codeforces Round #649 (Div. 2)D题[dfs求环+深度分层求图中独立集]

题目大意：

解题思路：

图论 ---- Codeforces Round #649 (Div. 2)D题[dfs求环+深度分层求图中独立集]相关推荐

最新文章

热门文章