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2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int n, m; // 点数,边数
int ind[N]; // in[N]:入度,所有入度为零的点,可以排在当前最前面的位置。
// 树和图的存储
int h[N], e[N], ne[N], idx;
void add(int a, int b) {
e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
vector<int> path;
// 拓扑
bool topsort() {
queue<int> q;
// 扫描所有入度为零的点入队列
for (int i = 1; i <= n; i++)
if (!ind[i]) {
q.push(i);
path.push_back(i);
}
while (q.size()) {
int u = q.front(); // 队列头
q.pop();
for (int i = h[u]; ~i; i = ne[i]) { // 遍历t的所有出边
int v = e[i];
if (--ind[v] == 0) { // 入度减1后是不是为0 (前序依赖为0)
q.push(v); // 为0则入队列
path.push_back(v);
}
}
}
// 如果一共n个结点进入过队列则表示存在拓扑序
return path.size() == n;
}
int main() {
// 初始化为-1
memset(h, -1, sizeof h);
cin >> n >> m;
for (int i = 0; i < m; i++) {
int a, b;
cin >> a >> b;
add(a, b);
ind[b]++; // 记录每个结点的入度
}
if (!topsort())
puts("-1");
else {
for (int i = 0; i < n; i++) printf("%d ", path[i]);
puts(""); // 有向无环图的拓扑序不唯一
}
return 0;
}