标签：Just joke 多校 Alice GG game connected graph nodes

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 GG with nn 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 GG and delete it from GG.

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

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 33 nodes and edge set {(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} are not.

输入描述:

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

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

1≤n≤1001≤n≤100

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

It's guaranteed that graph GG 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

复制

3 1
1 2

输出

复制

Bob

注意到要么删除的是一条边（m--）要么删除的是一棵树（n -= k, m -= (k - 1)），都会让点数和边数的和减少一个奇数。因此只需要判断(n + m)的奇偶性即可。

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n, m;
    cin >> n >> m;
    for(int i = 1; i <= m; i++) {
        int u, v;
        cin >> u >> v;
    }
    if((n + m) & 1) cout << "Alice";
    else cout << "Bob";
    return 0;
}

标签：Just,joke,多校,Alice,GG,game,connected,graph,nodes
来源： https://www.cnblogs.com/lipoicyclic/p/15062898.html

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。