搜索与图论-拓扑序列

为什么记录呢

因为不记录全忘了

虽然记了也不一定会看

1. 有向无环图一定有拓扑序列
2. 邮箱无环图 - 拓扑图

1. 入度为0的点作为起点
2. 入度为0的点入队列
3. 枚举出边 t->j
4. 删掉当前边，t->j . j的入度减1
5. 判断j的入度是否为0，来判断是否加入队列

1. 有环： 不存在入度为0的点

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>

using namespace std;

const int maxn = 100010;

int h[maxn], e[maxn], ne[maxn], idx;

int q[maxn],d[maxn];

int n;

int hh = 0, tt = -1;

void add(int a, int b){
    e[idx] = b;
    
    ne[idx] = h[a];
    
    h[a] = idx++;
}

bool topsort(){
    
    while(hh <= tt){
        int t = q[hh++];
        for(int i = h[t]; i != -1; i = ne[i]){
            int j = e[i];
            d[j]--;
            if(d[j] == 0){
                q[++tt] = j;
                // cout<<"j: "<< j << " "; 
            }
        }
    }
    // cout<<"tt " << tt << "n-1 "<< n-1 << '\n';
    return tt == n-1;
    
}

int main(){
    int m,a,b;
    
    memset(h , -1, sizeof h);
    cin >> n >> m;
    for(int i = 0; i < m; i++){
        cin>>a>>b;
        add(a,b);
        // cout<<"b  "<< b << " ";
        d[b]++;
        
    }
    
    
    for(int i = 1; i <= n; i++){
        if(d[i] == 0){
            // cout<<"i: " << i<<'\n';
            q[++tt] = i;
        }
    }
    
    if(topsort()){
        for(int i = 0; i < n; i++){
            cout<<q[i] << " ";
        }
        
    }else cout<<-1<< '\n';
    
    return 0;
}