Coderfroces 862 B . Mahmoud and Ehab and the bipartiteness

最新推荐文章于 2024-06-03 20:21:29 发布

转载最新推荐文章于 2024-06-03 20:21:29 发布 · 72 阅读

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原文链接：http://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7561973.html

本文探讨了在一个给定的树形结构中，如何通过添加边使其成为一个二分图，并在此基础上达到可添加边数的最大化。文章提供了一种算法实现思路及具体代码示例。

Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees.

A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (u, v) that belongs to the graph, u and v belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.

Dr. Evil gave Mahmoud and Ehab a tree consisting of n nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?

A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same .

Input

The first line of input contains an integer n — the number of nodes in the tree (1 ≤ n ≤ 10⁵).

The next n - 1 lines contain integers u and v (1 ≤ u, v ≤ n, u ≠ v) — the description of the edges of the tree.

It's guaranteed that the given graph is a tree.

Output

Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions.

Examples

Input

3
1 2
1 3

Output

Input

Output

Note

Tree definition: https://en.wikipedia.org/wiki/Tree_(graph_theory)

Bipartite graph definition: https://en.wikipedia.org/wiki/Bipartite_graph

In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0.

In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5).

一颗子树，将其变为二分图，最大可添加的边数。将所有结点标记为1或0，则二分图的最大边数为pos=ans₀*ans₁，所以答案就为pos-n+1；

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(true)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
vector<int>v[100005];
int vis[100005][2],x,y;
ll ans=0,n;
void dfs(int x)
{
    vis[x][0]=1;
    for(int i=0;i<v[x].size();i++)
    {
        if(vis[v[x][i]][0]) continue;
        vis[v[x][i]][1]=vis[x][1]^1;
        if(vis[v[x][i]][1]==1) ans++;
        dfs(v[x][i]);
    }
}
int main()
{
    scanf("%lld",&n);
    for(int i=0;i<n-1;i++)
    {
        scanf("%d%d",&x,&y);
        v[x].push_back(y);
        v[y].push_back(x);
    }
    memset(vis,0,sizeof(vis));
    dfs(1);
    printf("%lld\n",(n-ans)*ans-n+1);
    return 0;
}

转载于:https://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7561973.html