Tree2cycle
A tree with N nodes and N-1 edges is given. To connect or disconnect one edge, we need 1 unit of cost respectively. The nodes are labeled from 1 to N. Your job is to transform the tree to a cycle(without superfluous edges) using minimal cost.
A cycle of n nodes is defined as follows: (1)a graph with n nodes and n edges (2)the degree of every node is 2 (3) each node can reach every other node with these N edges.
但还是有些要注意的。
比如 dfs 深搜时,由于栈内存只有1M。所以要用
#pragma comment(linker,"/STACK:102400000,102400000")
这条语句,而且这种是c++编译器的加法,gcc不是这种。。
然后还要注意的是,边的内存是顶点数*2-2
代码
// // 4714.cpp // ACM_HDU // // Created by ipqhjjybj on 13-9-8. // Copyright (c) 2013年 ipqhjjybj. All rights reserved. // #pragma comment(linker,"/STACK:102400000,102400000") #include <cstdio> #include <iostream> using namespace std; const int N=1111111; int n; struct node{ int to,next; }Edge[N<<1]; int head[N],cnt; bool visit[N]; void addEdge(int u,int v){ Edge[cnt].to=v; Edge[cnt].next=head[u]; head[u]=cnt++; } int ans; int dfs(int u){ int sum=0; visit[u]=true; for(int q=head[u];q!=-1;q=Edge[q].next){ if(!visit[Edge[q].to]) sum+=dfs(Edge[q].to); } if(sum>1){ if(u==1) ans+=(sum-2)*2; else ans+=(sum-1)*2; return 0; } return 1; } int main(){ int t; scanf("%d",&t); while(t--){ scanf("%d",&n); cnt=0; memset(head,-1,sizeof(int)*(n+10)); memset(visit,false,sizeof(false)*(n+10)); for(int i=1,u,v;i<n;i++){ scanf("%d %d",&u,&v); addEdge(u,v); addEdge(v,u); } ans=0; dfs(1); printf("%d\n",ans+1); } return 0; }