2021牛客暑期多校训练营4 F. Just a joke（思维）

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;
}