C - Path Graph
https://atcoder.jp/contests/abc287/tasks/abc287_c
思路
判断所有点组成一个链的路径
根据每个节点的度来判断:
链路的度: 两边为节点度为1, 中间所有节点度为2.
1 -- 2 -- 2 -- ... -- 2 -- 1
同时要防止corner问题出现:
存在两个以上连通子图,但是这些子图不联通。
通过并查集方法判断是否存在多个连通子图。
https://zhuanlan.zhihu.com/p/93647900
参考:
https://www.cnblogs.com/Lanly/p/17071525.html
Code
https://atcoder.jp/contests/abc287/submissions/38448483
int n, m; map<int, int> degree; int find(vector<int>& fa, int x) { if(x == fa[x]) return x; else{ fa[x] = find(fa, fa[x]); //父节点设为根节点 return fa[x]; //返回父节点 } } int main() { cin >> n >> m; if (m != n-1){ cout << "No" << endl; return 0; } vector<int> fa(n+1); // initialize union set for(int i=1; i<=n; i++){ fa[i] = i; } for(int i=0; i<m; i++){ int s, t; cin >> s >> t; degree[s]++; degree[t]++; int sfa = find(fa, s); int tfa = find(fa, t); if (sfa != tfa){ fa[s] = tfa; } } map<int, int> cnt; for(int i=1; i<=n; i++){ int deg = degree[i]; if (deg >= 3 || deg==0){ cout << "No" << endl; return 0; } cnt[deg]++; } int one = cnt[1]; int two = cnt[2]; if(one != 2){ cout << "No" << endl; return 0; } if (two != n-2){ cout << "No" << endl; return 0; } int blocknum = 0; for(int i=1; i<=n; i++){ if(i == fa[i]){ blocknum++; } if(blocknum >= 2){ cout << "No" << endl; return 0; } } cout << "Yes" << endl; return 0; }
标签:int,Graph,--,find,fa,https,Path,节点 From: https://www.cnblogs.com/lightsong/p/17074109.html