LUOGU 1137

传送门

题目分析

拓扑排序：将图从度为0的点不断的剥掉外层的点，即可得到拓扑序，再按照拓扑序进行一遍简单的dp。

code

#include<bits/stdc++.h>
using namespace std;

const int N = 100000, M = 200000, OO = 0x3f3f3f3f;

int n, m;
int ecnt, adj[N + 5], go[M * 2 + 5], nxt[M * 2 + 5], in[N + 5], ans[N + 5], top[N + 5], _top;
bool vst[N + 5];
queue<int> que;
struct node{
	int from, to;
	node(){}
	node(int u, int v):from(u), to(v){}
}edges[M * 2 + 5];

inline void addEdge(int u, int v){
	nxt[++ecnt] = adj[u], adj[u] = ecnt, go[ecnt] = v;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(NULL), cout.tie(NULL);
	cin >> n >> m;
	for(int i = 1; i <= m; i++){
		int x, y;
		cin >> x >> y;
		addEdge(x, y);
		in[y]++;
	}
	for(int i = 1; i <= n; i++)
		if(in[i] == 0) que.push(i);
	while(!que.empty()){
		top[++_top] = que.front(); que.pop();
		for(int e = adj[top[_top]]; e; e = nxt[e]){
			int v = go[e];
			in[v]--;
			if(in[v] == 0) que.push(v);
		}
	}
	for(int i = 1; i <= n; i++) ans[i] = 1;
	for(int i = 1; i <= _top; i++){
		for(int e = adj[top[i]]; e; e = nxt[e])
			if(ans[go[e]] < ans[top[i]] + 1) ans[go[e]] = ans[top[i]] + 1;
	}
	for(int i = 1; i <= n; i++) cout << ans[i] << endl;
}