题解
很明显的二分图染色,
一个合法的连通块,必定是单链或者偶数环,所有有三条及以上的边肯定不合法,
不过点和边的数量太大了,dfs怕MLE,改成了bfs的
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+10;
const int mod=1e9+7;
vector<int>e[N];
int n,m,k;
int color[N];
ll bfs(){
ll ans=1;
queue<int>q;
memset(color, 0, sizeof(color));
for (int i = 1; i <= n; ++i) {
if(!color[i]&&!e[i].empty()){//单独的点直接忽略
ans=(ans*2)%mod;
q.push(i);
color[i]=1;
while(!q.empty()){
int u=q.front();
q.pop();
if(e[u].size()>2)return 0;
for (int j = 0; j <e[u].size(); ++j) {
int v=e[u][j];
if(!color[v]){
color[v]=3-color[u];
q.push(v);
}else if(color[v]==color[u])return 0;
}
}
}
}
return ans;
}
int main(){
ios::sync_with_stdio(0);
int T;
scanf("%d",&T);
for (int cs = 1,u,v; cs <= T; ++cs) {
scanf("%d%d",&n,&m);
for (int i = 1; i <= n; ++i) {
e[i].clear();
}
for (int i = 1; i <= m; ++i) {
scanf("%d%d",&u,&v);
e[u].push_back(v);
e[v].push_back(u);
}
printf("%lld\n", bfs());
}
return 0;
}