Codeforces 982 C Cut 'em all!(DFS)
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.
InputThe 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.
OutputOutput a single integer kk — the maximum number of edges that can be removed to leave all connected components with even size, or −
4 2 4 4 1 3 1output Copy
1input Copy
3 1 2 1 3output Copy
-1
10 7 1 8 4 8 10 4 7 6 5 9 3 3 5 2 10 2 5output Copy
4input Copy
2 1 2output Copy
0Note
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
.
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,l=0,L=0,x,y,ans=0;
int a[301000],link[301000],first[301000],ne[301000],father[301000],f[301000],p[301000];
void add(int x,int y)
{
a[++l]=x;link[l]=y;ne[l]=first[x];first[x]=l;
}
int build(int x,int fa)
{
for(int i=first[x];i!=-1;i=ne[i])
{
if(link[i]==fa) continue;
int c=build(link[i],x);
if(c%2==1) f[x]+=c;
else ans++;
}
return f[x]+1;
}
int main()
{
scanf("%d",&n);memset(first,-1,sizeof(first));
for(int i=1;i<=n-1;i++)
{
scanf("%d %d",&x,&y);
add(x,y);add(y,x);
}
if(n%2==1) {printf("-1");return 0;}
build(1,-1);
//for(int i=1;i<=n;i++) printf("%d\n",f[i]);
printf("%d",ans);
}