dfs找环 http://acm.hdu.edu.cn/showproblem.php?pid=6736 Forest Program
Forest Program
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 299 Accepted Submission(s): 111
Problem Description
The
kingdom of Z is fighting against desertification these years since
there are plenty of deserts in its wide and huge territory. The deserts
are too arid to have rainfall or human habitation, and the only
creatures that can live inside the deserts are the cactuses. In this
problem, a cactus in desert can be represented by a cactus in graph
theory.
In graph theory, a cactus is a connected undirected graph with no self-loops and no multi-edges, and each edge can only be in at most one simple cycle. While a tree in graph theory is a connected undirected acyclic graph. So here comes the idea: just remove some edges in these cactuses so that the remaining connected components all become trees. After that, the deserts will become forests, which can halt desertification fundamentally.
Now given an undirected graph with n vertices and m edges satisfying that all connected components are cactuses, you should determine the number of schemes to remove edges in the graph so that the remaining connected components are all trees. Print the answer modulo 998244353.
Two schemes are considered to be different if and only if the sets of removed edges in two schemes are different.
In graph theory, a cactus is a connected undirected graph with no self-loops and no multi-edges, and each edge can only be in at most one simple cycle. While a tree in graph theory is a connected undirected acyclic graph. So here comes the idea: just remove some edges in these cactuses so that the remaining connected components all become trees. After that, the deserts will become forests, which can halt desertification fundamentally.
Now given an undirected graph with n vertices and m edges satisfying that all connected components are cactuses, you should determine the number of schemes to remove edges in the graph so that the remaining connected components are all trees. Print the answer modulo 998244353.
Two schemes are considered to be different if and only if the sets of removed edges in two schemes are different.
Input
The
first line contains two non-negative integers n, m (1 ≤ n ≤ 300 000, 0 ≤
m ≤ 500 000), denoting the number of vertices and the number of edges
in the given graph.
Next m lines each contains two positive integers u, v (1 ≤ u, v ≤ n, u = v), denoting that vertices u and v are connected by an undirected edge.
It is guaranteed that each connected component in input graph is a cactus.
Next m lines each contains two positive integers u, v (1 ≤ u, v ≤ n, u = v), denoting that vertices u and v are connected by an undirected edge.
It is guaranteed that each connected component in input graph is a cactus.
Output
Output a single line containing a non-negative integer, denoting the answer modulo 998244353.
Sample Input
3 3
1 2
2 3
3 1
6 6
1 2
2 3
3 1
2 4
4 5
5 2
Sample Output
7
49
Source
Recommend
//#include <bits/stdc++.h> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <algorithm> #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdio.h> #include <queue> #include <stack> #include <map> #include <set> #include <string.h> #include <vector> #define ME(x , y) memset(x , y , sizeof(x)) #define SF(n) scanf("%d" , &n) #define rep(i , n) for(int i = 0 ; i < n ; i ++) #define INF 0x3f3f3f3f #define mod 998244353 #define PI acos(-1) using namespace std; typedef long long ll ; const int maxn = 505005 ; const int maxm = 2050000; ll n , m , sum , ans , vis[maxn] , dfn[maxn] , cnt; ll head[maxn]; struct Edge { ll to , next ; }e[maxm]; void add(ll u , ll v) { e[cnt].to = v ; e[cnt].next = head[u]; head[u] = cnt++; } void init() { memset(vis , 0 , sizeof(vis)); memset(dfn , 0 , sizeof(dfn)); memset(head , -1 , sizeof(head)); cnt = 0 , ans = 1 ; } ll qpow(ll base, ll n) { ll ans = 1; while(n) { if(n&1) ans=(ans%mod)*(base%mod)%mod; base = (base%mod) * (base%mod)%mod; n/=2; } return ans%mod; } void dfs(ll id , ll step , ll fa) { vis[id] = 1 , dfn[id] = step ; for(ll i = head[id] ; i != -1 ; i = e[i].next) { ll v = e[i].to ; //cout << i << " " << v << endl ; if(v == fa || vis[v] == 2) continue ; if(vis[v] == 1) { sum += step - dfn[v] + 1; ans *= (qpow(2 , step-dfn[v]+1)-1+mod) % mod ; ans %= mod ; } else { dfs(v , step+1 , id); } } vis[id] = 2 ; } int main() { scanf("%lld%lld" , &n , &m); init(); for(ll i = 1 ; i <= m ; i++) { ll u , v ; scanf("%lld%lld" , &u , &v); add(u , v); add(v , u); } for(ll i = 1 ; i <= n ; i++) { if(!vis[i]) dfs(i , 1 , -1); } ans *= qpow(2 , m - sum); ans %= mod ; printf("%lld " , ans); return 0; }