https://loj.ac/problem/6003

变化了下的最小路径覆盖，要注意边数巨多。。。

#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;int biao[100];const int maxn = 10009;int head[maxn];int cur[maxn];int tot = 0;struct edge{    int v,nex,w;}e[maxn*20];void addedge(int u,int v,int w){    e[tot] = (edge){v,head[u],w};    head[u] = tot++;    e[tot] = (edge){u,head[v],w};    head[v] = tot++;}int deep[maxn];int nowball;bool bfs(int S,int T){    for(int i=0;i<=2*nowball+1;i++){        deep[i] = 0;    }    queue
         
           q;    deep[S] = 1;    q.push(S);    while(!q.empty()){        int now = q.front();        q.pop();        for(int i=head[now];i!=-1;i=e[i].nex){            int v = e[i].v;            int w = e[i].w;            if(deep[v]!=0 || w<=0) continue;            deep[v] = deep[now]+1;            if(v==T) return deep[T];            q.push(v);        }    }    return deep[T];}int dfs(int now,int T,int maxflow){    if(now==T) return maxflow;    int all  =0;    for(int &i=cur[now];i!=-1 &&all 
          
           =nowball) continue; addedge(from*2,nowball*2+1,1); } addedge(0,nowball*2,1); addedge(nowball*2+1,1,1); kkk+=dinic(0,1); if(nowball-kkk<=n) continue; else{ break; } } nowball--; printf("%d\n",nowball); for(int i=1;i<=nowball;i++){ if(vis[2*i]==0) { print(2*i); printf("\n"); } } return 0;}