【题解】CF915D：Almost Acyclic Graph

原题传送门
如果枚举删边，然后拓扑判环，会小T一点
然后发现其实可以枚举哪个点入度-1，在判环，等价于删边
优化到了$O(n^2)$

Code：

#include <bits/stdc++.h>
#define maxn 500010
using namespace std;
struct Edge{
    
    
	int to, next;
}edge[maxn << 1];
int num, head[maxn], d[maxn], deg[maxn], n, m, flag;
queue <int> q;

inline int read(){
    
    
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

void addedge(int x, int y){
    
     edge[++num] = (Edge){
    
    y, head[x]}, head[x] = num; }

int main(){
    
    
	n = read(), m = read();
	for (int i = 1; i <= m; ++i){
    
    
		int x = read(), y = read();
		++d[y];
		addedge(x, y);
	}
	for (int k = 1; k <= n; ++k){
    
    
		for (int i = 1; i <= n; ++i) deg[i] = d[i];
		--deg[k];
	//	for (int i = 1; i <= n; ++i) printf("deg[%d]=%d, ", i, deg[i]);
		for (int i = 1; i <= n; ++i)
			if (!deg[i]) q.push(i);
		while (!q.empty()){
    
    
			int u = q.front(); q.pop();
			for (int i = head[u]; i; i = edge[i].next){
    
    
				int v = edge[i].to;
				--deg[v];
				if (deg[v] == 0) q.push(v);
			}
		}
		flag = 1;
		for (int i = 1; i <= n; ++i)
			if (deg[i] > 0){
    
     flag = 0; break; }
		if (flag) break;
	}
	if (flag) puts("YES"); else puts("NO");
	return 0;
}