Codeforces 982C Cut 'em all!

最新推荐文章于 2020-07-27 14:13:28 发布

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本文介绍了一道关于树形结构的算法题Cut'emall!，任务是在保证所有剩余连通部分节点数为偶数的前提下，求解可以删除的最大边数。文章分析了题目条件，并给出了实现代码。

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：给了n个顶点和n-1条边，保证给出的是一棵树，如果删除一条边能够使得剩下的连通部分均有偶数个点，则记录这条边，求最多能删除的边的个数

分析:如果给出的点的个数是奇数，无论怎么删除边都不会产生全是偶数的连通子树，输出-1。如果给出的点的个数是偶数，对于每一个结点u,父结点为v，u及其子树的结点总数为偶数的话就可以删除e(u,v)，因为偶数-偶数还是偶数，统计所有这样的结点个数即可。

代码:

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
vector<int>G[maxn];
int ans[maxn],vis[maxn];
int dfs(int x)
{
    for(int i = 0; i < G[x].size(); i++)
    {
        int u = G[x][i];
        if(!vis[u])
        {
            vis[u] = 1;
            ans[x] += dfs(u);
            vis[u] = 0;
        }
    }
    ans[x]++;
    return ans[x];
}
int main()
{
    int n;
    cin>>n;
    for(int i = 1; i <= n - 1; i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    vis[1] = 1;
    dfs(1);
//    for(int i = 1; i <= n; i++)
//        printf("%d ",ans[i]);
//    printf("\n");
    if(n % 2)
        printf("-1\n");
    else
    {
        int cnt = 0;
        for(int i = 1; i <= n; i++)
        {
            if(ans[i] % 2 == 0)
                cnt++;
        }
        cout<<cnt - 1<<endl;
    }

    return 0;
}