題面:https://www.luogu.org/problem/P5022

本題顯然是一個基環樹,然后由于一個基環上n條邊只會走n-1條邊,所以只要枚舉要刪的邊再求解即可。Code: #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<algorithm> #include<ctime> #include<vector> using namespace std; const int N=5005; struct Node{int to,next; }e[N<<2]; int head[N],cnt,n,m,u[N],v[N],visit[N],cutu,cutv,tot,now[N],ans[N]; bool vis[N]; vector<int> G[N]; int read(){int x=0,f=1;char ch=getchar();while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}return x*f; } void add(int u,int v){e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt; } void dfs(int x){if(vis[x]){return;}vis[x]=1;printf("%d ",x);int a[N],num=0;for(int i=1;i<=n;i++){a[i]=0;}for(register int i=head[x];i!=-1;i=e[i].next){a[++num]=e[i].to;}sort(a+1,a+1+num);for(int i=1;i<=num;i++){dfs(a[i]);} } void solve(int x){if(visit[x]){return;}visit[x]=1;now[++tot]=x;for(register int i=0;i<G[x].size();i++){int y=G[x][i];if((y==cutv&&x==cutu)||(x==cutv&&y==cutu)){continue;}solve(y);} } bool check(){for(register int i=1;i<=n;i++){if(ans[i]==now[i]){continue;}if(ans[i]>now[i]){return 1;}if(ans[i]<now[i]){return 0;}} } int main(){memset(head,-1,sizeof(head));n=read();m=read();for(register int i=1;i<=m;i++){u[i]=read();v[i]=read();add(u[i],v[i]);add(v[i],u[i]);G[u[i]].push_back(v[i]);G[v[i]].push_back(u[i]);}for(register int i=1;i<=n;i++){sort(G[i].begin(),G[i].end());}if(m==n-1){dfs(1);return 0;}else{for(register int i=1;i<=m;i++){tot=0,cutu=u[i],cutv=v[i];memset(visit,0,sizeof(visit));solve(1);if(tot==n){if(ans[1]==0){for(register int j=1;j<=n;j++){ans[j]=now[j];}}else if(check()){for(register int j=1;j<=n;j++){ans[j]=now[j];}}}}for(int i=1;i<=n;i++){printf("%d ",ans[i]);}return 0;} }

轉載于:https://www.cnblogs.com/ukcxrtjr/p/11382286.html

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