题目链接

题面:

题意:
选最少的简单链,覆盖树的所有的边,并输出每条链的两个端点。

官方题解:

代码:

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#include<bitset>
#include<map>
#include<unordered_map>
#include<set>
#define ui unsigned int
#define ll long long
#define llu unsigned ll
#define ld long double
#define pr make_pair
#define pb push_back
#define lc (cnt<<1)
#define rc (cnt<<1|1)
#define len(x) (t[(x)].r-t[(x)].l+1)
#define tmid ((l+r)>>1)
using namespace std;

const int inf=0x3f3f3f3f;
const ll lnf=0x3f3f3f3f3f3f3f3f;
const double dnf=1e18;
const int mod=998244353;
const double eps=1e-8;
const double pi=acos(-1.0);
const int hp=13331;
const int maxn=400100;
const int maxm=100100;
const int up=100000;

int head[maxn],ver[maxn],nt[maxn],tot=1;
int d[maxn];
vector<int>vc;

void add(int x,int y)
{
    ver[++tot]=y,nt[tot]=head[x],head[x]=tot;
}

void dfs(int x,int fa)
{
    if(d[x]==1)
    {
        vc.pb(x);
        return ;
    }
    for(int i=head[x];i;i=nt[i])
    {
        int y=ver[i];
        if(y==fa) continue;
        dfs(y,x);
    }
}

int main(void)
{
    int n,x,y;
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        scanf("%d%d",&x,&y);
        add(x,y),add(y,x);
        d[x]++,d[y]++;
    }
    if(n==1)
    {
        printf("1\n\n");
        return 0;
    }
    if(n==2)
    {
        printf("1\n1 2\n");
        return 0;
    }
    for(int i=1;i<=n;i++)
    {
        if(d[i]!=1)
        {
            dfs(i,0);
            break;
        }
    }
    int ans=(vc.size()+1)/2;
    printf("%d\n",ans);
    for(int i=0;i<ans;i++)
        printf("%d %d\n",vc[i],vc[(i+ans)%vc.size()]);
    return 0;
}