链接：https://ac.nowcoder.com/acm/contest/11255/F
来源：牛客网
 

题目描述

Alice and Bob are playing a game.

At the beginning, there is an undirected graph GGG with nnn nodes.

Alice and Bob take turns to operate, Alice will play first. The player who can't operate will lose the game.

Each turn, the player should do one of the following operations.

1. Select an edge of GGG and delete it from GGG.

2. Select a connected component of GGG which doesn't have any loop, then delete it from GGG.

Alice and Bob are smart enough, you need to find who will win this game.

A connected component of an undirected graph is a set of nodes such that each pair of nodes is connected by a path, and other nodes in the graph are not connected to the nodes in this set.

For example, for graph with 333 nodes and edge set {(1,2),(2,3),(1,3)}.{1,2,3}\{(1,2),(2,3),(1,3)\}. \{1,2,3\}{(1,2),(2,3),(1,3)}.{1,2,3} is a connected component but {1,2},{1,3}\{1,2\},\{1,3\}{1,2},{1,3} are not. 


 

输入描述:

The first line has two integers n,mn,mn,m.

Then there are mmm lines, each line has two integers (u,v)(u,v)(u,v) describe an edge in GGG.

1≤n≤1001\leq n\leq 1001≤n≤100

0≤m≤min(200,n(n−1)/2)0\leq m\leq min(200,n(n-1)/2)0≤m≤min(200,n(n−1)/2)

It's guaranteed that graph GGG doesn't have self loop and multiple edge.

输出描述:

Output the name of the player who will win the game.

示例1

输入

3 1 1 2

输出

Bob

题解

一开始就在想并查集的方法

后来学长加了一个环的后就出来了

#include<cstdio>
#include <map> 
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

inline int read() {int x=0,f=1;char c=getchar();while(c!='-'&&(c<'0'||c>'9'))c=getchar();if(c=='-')f=-1,c=getchar();while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();return f*x;}
typedef long long ll;
const int maxn=5e5+10;
int f[maxn];
int sum[maxn];
int ans;
struct node{
    int u,v;
}a[maxn];
int find(int x){
    if(f[x]==x){
        return x;
    }
    else{
        return f[x]=find(f[x]);
    }
}
void unio(int x,int y){
    int f1=find(x);
    int f2=find(y);
    if(f1!=f2){
        f[f1]=f2;
    }
    else
        ans++;
}
int main(){
    int n,m;
	int k=0;
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        f[i]=i;
    }
    ans=0;
    for(int i=0;i<m;i++){
        cin>>a[i].u>>a[i].v;
        sum[a[i].u]++;
        sum[a[i].v]++;
    }
    for(int i=0;i<m;i++){
    	unio(a[i].u,a[i].v);
    	}
    for(int i=1;i<=n;i++)
    {
    	if(f[i]==i)
    	k++;
	}
	cout<<u<<" "<<k<<endl;
	if((u+k)%2==0)
	cout<<"Bob"<<endl;
    else
    cout<<"Alice"<<endl;
}

看完题解，发现只要判断n+m的奇数偶数就可以了 

没发现，哎。

奇偶判断代码就不贴了