1492.Problem_A
Time Limit: 1000 MS Memory Limit: 32768 KB
Description SDNU ACM-ICPC Association is in a mess. If you join it, you should be in the shadow of a person. This person, we call him "dalao". For example, there are two rookies, C and D, rely on A. We can say that A is the closest dalao of rookies C and D.For another example, A have two heelers C and D, D have two heelers E and F.We can know that A is the closest dalao of E and C.
But with the growth of Association, we cannot find the closest dalao between two rookies. So we need your help.
Input There will be a series of examples. For each input, the first line contains two integers N. N means the number of people. Followed by the N-1 lines, each line contains two integers A and B. It means that A is a dalao of B. Then the next line, contains two integers A and B. It means that we want to know who is the closest dalao between A and B. Output
For each input, you should output the name of the closest dalao.If you cannot find this dalao, you should output "I am so bad."
Sample Input 7 1 2 1 3 2 5 2 6 6 7 6 8 3 5 Sample Output 1 Hint Remember that a rookie is also a dalao of itself For example: Input 3 1 2 2 3 2 3 Output 2
选拔赛的一道题目，题意很好理解，简单说就是LCA（最近公共祖先），不过之前没怎么做过这类题目，所以也一直没有整理过自己的模板.....翻到一个dfs+ST的算法就直接敲了，然后....TLE了，才看到后面有更快的算法，不过离比赛结束还剩10分钟已经没时间了，所以就这样放弃了。结果比赛结束下来，敲的后面的倍增算法就A了.....早知道该先敲后面的说，哎呀好气啊。
下面是AC代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<queue> using namespace std; const int MAXN=10010; const int DEG=20; struct Edge { int to,next; }edge[MAXN*2]; int head[MAXN],tot; void addedge(int u,int v) { edge[tot].to=v; edge[tot].next=head[u]; head[u]=tot++; } void init() { tot=0; memset(head,-1,sizeof(head)); } int fa[MAXN][DEG]; int deg[MAXN]; void BFS(int root) { queue<int>que; deg[root]=0; fa[root][0]=root; que.push(root); while(!que.empty()) { int tmp=que.front(); que.pop(); for(int i=1;i<DEG;i++) { fa[tmp][i]=fa[fa[tmp][i-1]][i-1]; } for(int i=head[tmp];i!=-1;i=edge[i].next) { int v=edge[i].to; if(v==fa[tmp][0]) continue; deg[v]=deg[tmp]+1; fa[v][0]=tmp; que.push(v); } } } int LCA(int u,int v) { if(deg[u]>deg[v]) swap(u,v); int hu=deg[u],hv=deg[v]; int tu=u,tv=v; for(int det=hv-hu,i=0;det;det>>=1,i++) { if(det&1) tv=fa[tv][i]; } if(tu==tv) return tu; for(int i=DEG-1;i>=0;i--) { if(fa[tu][i]==fa[tv][i]) continue; tu=fa[tu][i]; tv=fa[tv][i]; } return fa[tu][0]; } bool flag[MAXN]; int main() { int T; int n; int u,v; while(scanf("%d",&n)!=EOF) { init(); memset(fa,0,sizeof(fa)); memset(deg,0,sizeof(deg)); memset(flag,false,sizeof(flag)); for(int i=1;i<n;i++) { scanf("%d%d",&u,&v); addedge(u,v); addedge(v,u); flag[v]=true; } int root; for(int i=1;i<=n;i++) { if(!flag[i]) { root=i; break; } } BFS(root); scanf("%d%d",&u,&v); if(LCA(u,v)==0) cout<<"I am so bad."<<endl; else cout<<LCA(u,v)<<endl; } return 0; }