Cut 'em all! CodeForces - 982C（dfs）

最新推荐文章于 2020-07-27 14:13:28 发布

原创最新推荐文章于 2020-07-27 14:13:28 发布 · 350 阅读

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本文探讨了一种针对树形结构的算法问题，即通过合理切割边的方式将树分割成多个节点数量为偶数的子树，以实现最大数量的边切割。文章详细介绍了算法的实现思路，并提供了一个具体的C++代码示例。

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You're given a tree with $n$ 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 $n$ ( $1 \leq n \leq 10^{5}$ ) denoting the size of the tree.

The next $n - 1$ lines contain two integers $u$ , $v$ ( $1 \leq u, v \leq n$ ) each, describing the vertices connected by the $i$ -th edge.

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

Output

Output a single integer $k$ — the maximum number of edges that can be removed to leave all connected components with even size, or $- 1$ if it is impossible to remove edges in order to satisfy this property.

Examples
inputCopy
4
2 4
4 1
3 1
outputCopy
1
inputCopy
3
1 2
1 3
outputCopy
-1
inputCopy
10
7 1
8 4
8 10
4 7
6 5
9 3
3 5
2 10
2 5
outputCopy
4
inputCopy
2
1 2
outputCopy
0

Note

In the first example you can remove the edge between vertices $1$ and $4$ . 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 $- 1$

题意：有一颗树，现在，你需要切掉一些边，使这颗树分成若干个节点为偶数的部分，问，最多能切掉几条边。不能切,输出-1。

思路：首先，只有节点数为偶数的才能分成若干个偶数的部分，所以节点数为奇数的就直接输出-1吧。如果是偶数的话，从任意一点进行dfs，dfs返回的是子树节点的个数，如果是偶数的话，那么，父节点与子树的这条边一定是可以切掉的，所以，只要记录子树节点为偶数的个数就可以了。

#include "iostream"
#include "vector"
#include "cstdio"
using namespace std;
const int Max=1e5+10;
vector<int > G[Max];
bool flag[Max];//避免重复计算
int ans;
int dfs(int x)
{
    int res=0;
    for(int i=0;i<G[x].size();i++){
        int v=G[x][i];
        if(!flag[v]){
            flag[v]=1;
            int tmp=dfs(v);
            if(tmp%2==0) ans++;//如果子树节点的个数是偶数，那么，这条边就可以切掉
            res+=tmp;
        }
    }
    return res+1;//子树节点加一，就是这颗树的节点数
}
int main()
{
    int n,u,v;
    ans=0;
    scanf("%d",&n);
    for(int i=0;i<n-1;i++){
        scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    if(n%2!=0)//如果节点数为奇数，那么一定不可以切边
        printf("-1\n");
    else{
        flag[1]=1;
        dfs(1);
        printf("%d\n",ans);
    }
    return 0;
}