题意

题目链接

Sol

最直观的思路是求出删除每个点后的最长路，我们考虑这玩意儿怎么求

设\(f[i]\)表示以\(i\)结尾的最长路长度，\(g[i]\)表示以\(i\)开始的最长路长度

根据DAG的性质，显然我们删除一个点后，整个集合会被分成两部分：拓扑序小于/大于当前点

那么此时的最长路一定可以通过计算连接着两个集合的边\((u, v)\)的\(f(u) + f(v) +1\)得到

这样的话我们可以直接维护边集，在统计每个点的答案的时候首先删掉入边的贡献统计答案，统计完后再加入出边的贡献

显然线段树可以维护，其实堆也可以维护，具体见代码(抄袭自yyb大佬)

#include<bits/stdc++.h>
#define chmax(x, y) (x = (x > y ? x : y))
#define chmin(x, y) (x = (x < y ? x : y))
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
inline int read() {char c = getchar(); int x = 0, f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f;
}
int N, M, a1 = INF, a2;
class MyPriorityQueue {public: priority_queue<int> q1, q2;void push(int x) {q1.push(x);}int pop(int x) {q2.push(x);}bool empty() {while(!q2.empty() && (q1.top() == q2.top())) q1.pop(), q2.pop();return q1.size() == 0;}int top() {return empty() ? INF : q1.top();}
};
MyPriorityQueue Q;
struct Graph {vector<int> v[MAXN];int f[MAXN], inder[MAXN], id[MAXN], tot;Graph() {tot = 0;}void AddEdge(int x, int y) {v[x].push_back(y); inder[y]++;}void Topsort() {queue<int> q;for(int i = 1; i <= N; i++) if(!inder[i]) q.push(i);while(!q.empty()) {int p = q.front(); q.pop(); id[++tot] = p;for(int i = 0; i < v[p].size(); i++) {int to = v[p][i]; chmax(f[to], f[p] + 1);if(!(--inder[to])) q.push(to);}}}
};
Graph Gs, Gt;
int main() {N = read(); M = read();for(int i = 1; i <= M; i++) {int x = read(), y = read();Gs.AddEdge(x, y); Gt.AddEdge(y, x);}Gs.Topsort(); Gt.Topsort();for(int i = 1; i <= N; i++) Q.push(Gt.f[i]);for(int t = 1; t <= N; t++) {int x = Gs.id[t]; Q.pop(Gt.f[x]);for(int i = 0; i < Gt.v[x].size(); i++) {int to = Gt.v[x][i];Q.pop(Gs.f[to] + Gt.f[x] + 1);}int now = Q.top(); Q.push(Gs.f[x]);if(now < a1) a1 = now, a2 = x;for(int i = 0; i < Gs.v[x].size(); i++) {int to = Gs.v[x][i];Q.push(Gs.f[x] + Gt.f[to] + 1);}}printf("%d %d\n", a2, a1);return 0;
}