最短路樹就是用bfs走一遍就可以了 d[v] = d[u] + 1 表示v是u的前驅邊

然後遍歷每個結點 存下它的前驅邊 再用dfs遍歷每個結點 依次取每個結點的某個前驅邊即可

#include <bits/stdc++.h>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 1e6+10, INF = 0x7fffffff;

int n, m, k, cnt;
int head[maxn], d[maxn];
vector<int> f[maxn];
vector
 
<string> g;
char str[maxn];
struct node
{
    int u, v, next;
}Node[maxn<<1];

void add_(int u, int v)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].next = head[u];
    head[u] = cnt++;
}

void add(int u, int v)
{
    add_(u, v);
    add_(v, u);
}

void init()
{
    mem(head, 
 
-1);
    cnt = 0;
}

void dfs(int u)
{
    if(g.size() >= k) return;
    if(u == n+1) { g.push_back(str); return; }
    //cout<< 111 <<endl;
    for(int i=0; i<f[u].size(); i++)
    {
        str[f[u][i]/2] = ‘1‘;
       // cout<< str <<endl;
        dfs(u+1);
        str[f[u][i]
 
/2] = ‘0‘;
    }
}

void bfs(int u)
{
    mem(d, -1);
    queue<int> Q;
    Q.push(u);
    d[u] = 0;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i=head[u]; i!=-1; i=Node[i].next)
        {
            node e = Node[i];
            if(d[e.v] == -1)
            {
                d[e.v] = d[u] + 1;
                Q.push(e.v);
            }
        }
    }
}

int main()
{
    init();
    int u, v;
    cin>> n >> m >> k;
    for(int i=0; i<m; i++)
    {
        cin>> u >> v;
        add(u, v);
    }
    bfs(1);
   // cout<< 11 <<endl;
    for(int i=1; i<=n; i++)
    {
        for(int j=head[i]; j!=-1; j=Node[j].next)
            if(d[Node[j].v] + 1 == d[i])
            {
                f[i].push_back(j);
               // cout<< i << "  " << j/2 <<endl;
            }
    }
    for(int i=0; i<m; i++) str[i] = ‘0‘;
    dfs(2);
    cout<< g.size() <<endl;
    for(int i=0; i<g.size(); i++)
    {
        cout<< g[i] <<endl;
    }

    return 0;
}

Berland and the Shortest Paths CodeForces - 1005F（最短路樹）