codeforces 982C （dfs）

题意描述

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.

给定一棵树，询问最多删除多少条边可以使每个连通块的数量都是偶数

思路

这道题的even最初理解成了相同的意思，导致没读懂题。要想最多删除边，肯定要使每个连通块的数量都为2。如果结点数是奇数，则不满足题意，输出-1.否则就dfs寻找所有结点的子结点数，如果子结点数量为偶数，则说明可以删除一条边。

AC代码

#include<bits/stdc++.h>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
const int N=2*1e5+10;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
vector<int> g[N];
int n;
int ans=0;
bool used[N];
int dfs(int u){
    int res=0;
    used[u]=true;
    rep(i,0,g[u].size()){
        int v=g[u][i];
        if(!used[v]){
            int cur=dfs(v);
            if(cur%2==0) ans++;
            res+=cur;
        }
    }
    return res+1;
}
void solve(){
    cin>>n;
    rep(i,0,n-1){
        int x,y;cin>>x>>y;
        g[x].PB(y);
        g[y].PB(x);
    }
    if(n%2!=0){
        cout<<-1<<endl;
        return;
    }
    dfs(1);
    cout<<ans<<endl;
}
int main(){
    IOS;
    solve();
    return 0;
}

本文地址：https://blog.csdn.net/weixin_45729946/article/details/107610660