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$\lfloor\frac{m}{2}\rfloor$, $dfs$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 '\n' #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\n",cnt); REP(i,1,cnt) printf("%d %d %d\n",f[i].x,f[i].y,f[i].z); }