码风有亿点点抽象

#include <stdio.h>
#include <stdlib.h>
#include <vector>

#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))

long long read(){
	char ch=getchar();
	long long res=0;
	short f=1;
	for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f*=-1;
	for(;ch>='0'&&ch<='9';ch=getchar())res=(res<<3)+(res<<1)+(ch^48);
	return res*f;
}

class Solution{
	
	public:
		
		void solve(){
			for(int i = 1; i <= n; i++) if(order[i] == 0) Tarjan(i, 0);
			int cnt = 0, index = 0, ans = 0;
			for(int i = 1; i <= n; i++) if(top[i] != index && degree[top[i]] == 0) index = top[i], cnt++;
			if(cnt == 1) for(int i = 1; i <= n; i++) if(top[i] == index) ans++;
			printf("%d\n", ans);
			return ;
		}
		
		void print(){
			printf("order :\n");
			for(int i = 1; i <= n; i++)
				printf("%d ", order[i]);
			putchar('\n');
			printf("top :\n");
			for(int i = 1; i <= n; i++)
				printf("%d ", top[i]);
			putchar('\n');
			printf("degree :\n");
			for(int i = 1; i <= n; i++)
				printf("%d ", degree[i]);
			putchar('\n');
			return ;
		}
		
		Solution() : n(0), m(0){}
		Solution(int __n, int __m) : n(__n), m(__m), path(__n + 1), order(__n + 1), top(__n + 1), degree(__n + 1){
			for(int i = 0; i < m; i++){
				int u = read(), v = read();
				path[u].emplace_back(v);
				path[v].emplace_back(u);
			}
		}
		
	protected:
		
		int n, m;
		int count = 0;
		
		std::vector<std::pair<int, int>>ans;
		std::vector<std::vector<int>>path;
		std::vector<int>order, top, degree;
		
		void Tarjan(int u, int pre){
			top[u] = order[u] = ++count;
			for(int i = 0; i < path[u].size(); i++){
				int v = path[u][i];
				if(order[v] != 0 && v != pre) top[u] = std::min(top[u], order[v]);
				if(order[v] == 0){
					Tarjan(v, u);
					if(order[u] < top[v]) degree[order[u]]++;
					top[u] = std::min(top[u], top[v]);
				}
			}
			return ;
		}
		
};

int main(){
	
	int n = read(), m = read();
	Solution ans(n, m);
	ans.solve();
	//ans.print();
	
	return 0;
}

记录详情 ，可以看到很多是因为输出0而WA

但更抽象的是，我最开始没看题建的是双向边，然后是40pts，但注意到这个问题后改成建单边后反倒变成28pts了