大意: 给定$n$节点$m$条边无向图, 不保证连通, 求选出最多邻接边, 每条边最多选一次.
上界为$lfloorfrac{m}{2} floor$, $dfs$贪心划分显然可以达到上界.
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif
int n, m, cnt, vis[N];
vector<int> g[N];
struct {int x,y,z;} f[N];
void add(int x, int y, int z) {
f[++cnt]={x,y,z};
}
int dfs(int x) {
vis[x] = ++*vis;
int lst = 0;
for (int y:g[x]) {
if (!vis[y]) {
int t = dfs(y);
if (t) add(x,y,t);
else if (lst) add(lst,x,y),lst=0;
else lst = y;
}
else if (vis[y]>vis[x]) {
if (lst) add(lst,x,y),lst=0;
else lst = y;
}
}
return lst;
}
int main() {
scanf("%d%d", &n, &m);
REP(i,1,m) {
int u=rd(),v=rd();
g[u].pb(v),g[v].pb(u);
}
REP(i,1,n) if (!vis[i]) dfs(i);
printf("%d
",cnt);
REP(i,1,cnt) printf("%d %d %d
",f[i].x,f[i].y,f[i].z);
}