Codeforces 982 C Cut 'em all!(DFS)

C. Cut 'em all!
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

Input

The first line contains an integer nn (1n1051≤n≤105) denoting the size of the tree.

The next n1n−1 lines contain two integers uuvv (1u,vn1≤u,v≤n) each, describing the vertices connected by the ii-th edge.

It's guaranteed that the given edges form a tree.

Output

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.

Examples
input
Copy
4
2 4
4 1
3 1
output
Copy
1
input
Copy
3
1 2
1 3
output
Copy
-1
input
Copy
10
7 1
8 4
8 10
4 7
6 5
9 3
3 5
2 10
2 5
output
Copy
4
input
Copy
2
1 2
output
Copy
0
Note

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);
}

posted @ 2018-07-07 14:44  The-Pines-of-Star  阅读(169)  评论(0编辑  收藏  举报