AGC017D Game on Tree [博弈论]

Description:
树上删边游戏。

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Solution:
具体见程序。

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#include 
using namespace std;
const int maxn = 1e5 + 5;
int n;
int sg[maxn];
vector<int> G[maxn]; 
void dfs(int u, int last) {
    for(int i = 0; i < G[u].size(); ++i) {
        int v = G[u][i];
        if(v == last) {
            continue;
        }
        dfs(v, u);
        sg[u] ^= sg[v] + 1;
    }
}
int main() {
    scanf("%d", &n);
    for(int i = 1; i < n; ++i) {
        int u, v;
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    dfs(1, 0);
    puts(sg[1] ? "Alice" : "Bob");
    return 0;
}