2019牛客多校 H Stammering Chemists (模拟)

这题找到他们的不同特征判断就好了
题意还说 不是下面的 随便输出 就可以少盘一种了 虽然也没有少写啥
第一个 连边 只有2个是出现1次的
4 和 5 用 2个 3边 和 一个4边 判断

2 和 3 我dfs2边 3 的话 4深度出现2次 剩下的直接出 2图就好

#include <bits/stdc++.h>
#define int long long
using namespace std;
 
int f[100];
int n, m;
 
string str = "*COFFEECHICKEN";
 
int head[10], cnt;
int nxt[15], to[15];
 
void ade(int a, int b) {
    to[++cnt] = b;
    nxt[cnt] = head[a], head[a] = cnt;
}
 
int d[15];
void dfs(int x, int fa, int dep){
    d[x] = dep;
    for(int i = head[x]; i; i = nxt[i]) {
        if(to[i] == fa) continue;
        dfs(to[i], x, dep + 1);
    }
}
int du[15];
signed main() {
    int cas;
    cin >> cas;
    while(cas --) {
        memset(head, 0, sizeof head);
        cnt = 0;
        memset(du, 0, sizeof du);
        int du1 = 0, du2 = 0, du3 = 0, du4 = 0;
        for(int i = 1, a, b; i <= 5; i ++) {
            cin >> a >> b;
            ade(a, b), ade(b, a);
            du[a] ++, du[b] ++;
        }
         
        for(int i = 1; i <= 6; i ++) {
            if(du[i] == 1) du1 ++;
            if(du[i] == 2) du2 ++;
            if(du[i] == 3) du3 ++;
            if(du[i] == 4) du4 ++;
        }
    // cout << du1 << " " << du2 << " " << du3 << " " << du4 << endl;
        if(du4 != 0) {
            cout << "2,2-dimethylbutane" << endl;
        }else if(du3 == 2) {
            cout << "2,3-dimethylbutane" << endl;
        }else if(du1 == 2 && du2 == 4) {
            cout << "n-hexane" << endl;
        }else {
            memset(d, 0, sizeof d);
            int rt = 1;
            dfs(rt, 0, 1);
            int tmp = 0;
            for(int i = 1; i <= 6; i ++) {
            // cout << d[i] << " ";
                if(d[i] > tmp) rt = i, tmp = d[i];
            }
            memset(d, 0, sizeof d);
            dfs(rt, 0, 1);
            int d4 = 0, d5 = 0;
            for(int i = 1; i <= 6; i ++) {
            // cout << d[i] << " ";
                if(d[i] == 4) d4 ++;
            }
            //cout << d4 << endl;
            if(d4 == 2) {
                cout << "3-methylpentane" << endl;
            }else {
                cout << "2-methylpentane" << endl;
            }
        }
         
    }
    return 0;
}