Educational Codeforces Round 56 (Rated for Div. 2) D

给你一个无向图 以及点的个数和边  每个节点只能用1 2 3 三个数字 求相邻

两个节点和为奇数   能否构成以及有多少种构成方法

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define maxn 300005
const LL mod = 998244353;
vector<int>q[maxn];
LL fa[maxn],sumx,sumy;
bool flag = 0;
LL poww(LL a,LL b){
    LL ans = 1;
    while(b){
        if(b%2){
            ans = ans*a%mod;
        }
        b/=2;
        a = a*a%mod;
    }
    return ans%mod;
}
void dfs(int u,int deep){
    if(flag||q[u].size()==0) return ;
    for(int j=0;j<q[u].size();j++){
        int v = q[u][j];
        if(fa[v]==-1){
            fa[v] = 1-deep;
            sumx++;
            if(deep==1) sumy++;
            dfs(v,1-deep);
        }else{
            if((1-deep)!=fa[v]){
                flag = 1;
                return ;
            }
        }
    }
}
int main(){
    int t;
    cin>>t;
    while(t--){
        flag = 0;
        int n,m;
        scanf("%d%d",&n,&m);
        LL ans = 1;
        vector<int>s,z;
        for(int j=0;j<=n;j++){
            q[j].clear();
            fa[j]=-1;
        }
        for(int j=0;j<m;j++){
            int u,v;
            scanf("%d%d",&u,&v);
            q[u].push_back(v);
            q[v].push_back(u);
        }
        for(int j=1;j<=n;j++){
            sumx = 1,sumy = 1;
            if(fa[j]==-1){
                fa[j]=0;
                dfs(j,0);
                ans = (ans*(poww(1LL*2,1LL*sumy)+poww(1LL*2,1LL*(sumx-sumy)))%mod)%mod;
            }
        }
        if(flag){
            ans = 0;
        }
        cout<<ans<<endl;

    }
}