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对每一段都有两种情况，选和不选，二进制枚举每一种情况（或则dfs），然后判断选择的段会组成几个连通块，当连通块的数量为1时答案+1 code： #include<iostream> #include<cstring> using namespace std; const int N = 100; int head[N],e[N],ne[N]; int cnt; int ans; bool st[10]; int fa[10]; void add(int u,int.

题目 C - Skill Up Time Limit: 2 sec / Memory Limit: 1024 MB Score :300300points Problem Takahashi, who is a novice in competitive programming, wants to learnMMalgorithms. Initially, hisunderstanding levelof each of theMMalgorithms is00. Takaha...