数据范围过小怎么做都行。考虑优秀一点的做法。考虑dfs树上两台中心服务器间的路径,路径上所有能割掉中心服务器所在子树的点均可以成为答案。直接从两点中的任意一点开始dfs就更方便了。一开始弱智的以为只要是路径上的割点都能作为答案,wa了无数发。当然建棵圆方树也完全没问题。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 110 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,a[N][N],deep[N],dfn[N],size[N],low[N],fa[N],cnt,S,T,ans; void tarjan(int k) { dfn[k]=low[k]=++cnt;size[k]=1; for (int i=1;i<=n;i++) if (a[k][i]) { if (!dfn[i]) { tarjan(i); fa[i]=k;size[k]+=size[i]; low[k]=min(low[k],low[i]); if (dfn[i]<=dfn[T]&&dfn[i]+size[i]-1>=dfn[T]&&low[i]>=dfn[k]&&k!=S&&k!=T) ans=min(ans,k); } else low[k]=min(low[k],dfn[i]); } } int main() { #ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); const char LL[]="%I64d\n"; #else const char LL[]="%lld\n"; #endif n=read(); S=read(),T=read(); while (S) { a[S][T]=a[T][S]=1; S=read(),T=read(); } S=read(),T=read(); ans=n+1;tarjan(S); if (ans>n) cout<<"No solution"; else cout<<ans; return 0; }