You're given a tree with nn vertices.
Your task is to determine the maximum possible number of edges that can be removed in such a way that all the remaining connected components will have even size.
The first line contains an integer nn (1≤n≤1051≤n≤105) denoting the size of the tree.
The next n−1n−1 lines contain two integers uu, vv (1≤u,v≤n1≤u,v≤n) each, describing the vertices connected by the ii-th edge.
It's guaranteed that the given edges form a tree.
Output a single integer kk — the maximum number of edges that can be removed to leave all connected components with even size, or −1−1if it is impossible to remove edges in order to satisfy this property.
4 2 4 4 1 3 1
1
3 1 2 1 3
-1
10 7 1 8 4 8 10 4 7 6 5 9 3 3 5 2 10 2 5
4
2 1 2
0
In the first example you can remove the edge between vertices 11 and 44. The graph after that will have two connected components with two vertices in each.
In the second example you can't remove edges in such a way that all components have even number of vertices, so the answer is −1−1.
题面很简单,就是说给你一棵树,让你删掉一些边之后使得所有连通块都是偶数个点。可以的话输出最多能删的边数,否则输出-1;
做法也比较简单,随便选一个点当做根,dfs数一下当前点的子孙节数(包括自己)如果是偶数,说明当前点和当前的父节点之间的边是可以删除的,ans+1,去掉根节点的那次即可。然后如果节点数本身就是奇数个,那当然是不可能有满足题意的情况的,特判一下。
下面是代码:
#include <bits/stdc++.h> #include <cstring> using namespace std; int ans=0; vector<int> vex[100005]; int dfs(int pre,int now) { int sum=1; for(int i=0;i<vex[now].size();i++) { if(vex[now][i]!=pre) sum+=dfs(now,vex[now][i]); } //cout<<pre<<' '<<now<<' '<<sum<<endl; if(sum%2==0) ans++; return sum; } int main() { int n; scanf("%d",&n); int a,b; for(int i=0;i<n-1;i++) { scanf("%d%d",&a,&b); vex[a].push_back(b); vex[b].push_back(a); } if(n&1) { printf("-1\n"); return 0; } dfs(0,1); printf("%d\n",ans-1); return 0; }