题解
推结论大题……然而我推不出什么结论
奇环显然是NO
如果一个联通块里有两个分离的环，也是NO
如果一个联通块里，点数为n，边数为m
m <= n的时候，是YES
m >= n + 2的时候，肯定是NO
m = n + 1的时候
如果只存在一个双联通分量的话，取出这个双联通分量，存在两个点点度为3
他们之间有3条路径，如果其中两条为2的话答案是YES，否则是NO
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define MAXN 20005
#define eps 1e-8
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
struct node {
int to,next;
}E[MAXN * 2];
int N,M,head[MAXN],sumE,col[MAXN];
int dfn[MAXN],low[MAXN],idx,sta[MAXN],top,cnt,Ecnt,Ncnt,tot;
int vis[MAXN],tims,deg[MAXN],path[5],pc;
vector<int> ver[MAXN];
void add(int u,int v) {
E[++sumE].to = v;
E[sumE].next = head[u];
head[u] = sumE;
}
void Init() {
read(N);read(M);
sumE = 0;memset(head,0,sizeof(head));
memset(col,0,sizeof(col));memset(vis,0,sizeof(vis));tims = 0;
memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));idx = 0;
int u,v;
for(int i = 1 ; i <= M ; ++i) {
read(u);read(v);
add(u,v);add(v,u);
}
}
bool color(int u) {
if(!col[u]) col[u] = 2;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(!col[v]) {
col[v] = col[u] ^ 1;
if(!color(v)) return false;
}
else if(col[v] == col[u]) return false;
}
return true;
}
void Tarjan(int u) {
dfn[u] = low[u] = ++idx;
sta[++top] = u;
++Ncnt;
for(int i = head[u] ; i ; i = E[i].next) {
++Ecnt;
int v = E[i].to;
if(dfn[v]) {low[u] = min(low[u],dfn[v]);}
else {
Tarjan(v);
if(low[v] >= dfn[u]) {
int t = 1;
++tot;ver[tot].clear();ver[tot].pb(u);
while(1) {
int x = sta[top--];
++t;
ver[tot].pb(x);
if(x == v) break;
}
if(t >= 4) ++cnt;
}
low[u] = min(low[u],low[v]);
}
}
}
void dfs_path(int u,int fa,int ed,int dep) {
if(u == ed) {path[++pc] = dep;return;}
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(vis[v] == tims && v != fa) {
dfs_path(v,u,ed,dep + 1);
}
}
}
void Solve() {
for(int i = 1 ; i <= N ; ++i) {
if(!col[i]) {
if(!color(i)) {puts("NO");return;}
}
}
for(int i = 1 ; i <= N ; ++i) {
if(!dfn[i]) {
top = 0;cnt = 0;Ecnt = 0;Ncnt = 0;tot = 0;
Tarjan(i);
Ecnt /= 2;
if(Ecnt <= Ncnt) continue;
if(Ecnt >= Ncnt + 2) {puts("NO");return;}
if(cnt >= 2) {puts("NO");return;}
++tims;
int st,ed;
for(int j = 1 ; j <= tot; ++j) {
if(ver[j].size() >= 4) {
for(int k = 0 ; k < ver[j].size() ; ++k) {
vis[ver[j][k]] = tims;
deg[ver[j][k]] = 0;
}
for(int k = 0 ; k < ver[j].size() ; ++k) {
int u = ver[j][k];
for(int h = head[u] ; h ; h = E[h].next) {
int v = E[h].to;
if(vis[v] == tims) {
++deg[u];
}
}
}
st = 0;ed = 0;
for(int k = 0 ; k < ver[j].size() ; ++k) {
int u = ver[j][k];
if(deg[u] == 3) {
if(!st) st = u;
else if(!ed) ed = u;
}
}
break;
}
}
pc = 0;
dfs_path(st,0,ed,0);
sort(path + 1,path + pc + 1);
if(path[1] == 2 && path[2] == 2) continue;
else {puts("NO");return;}
}
}
puts("YES");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
int T;
read(T);
while(T--) {
Init();
Solve();
}
return 0;
}