53 lines
1.3 KiB
C++
53 lines
1.3 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 110, M = N * N;
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int n;
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int din[N];
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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vector<int> path;
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// 求拓扑序的模板
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/*
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① 所有入度为0的点入队列
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② 队头元素出队，遍历它相临的点，并且将相临的点入度-1
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③ 如果减1后的入度为0，则此点入队列，意味着它的前序依赖已完成
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*/
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void topsort() {
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queue<int> q;
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// 所有入度为0的入队列
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for (int i = 1; i <= n; i++)
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if (din[i] == 0) q.push(i);
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while (q.size()) {
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int u = q.front();
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q.pop();
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path.push_back(u);
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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din[v]--;
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if (din[v] == 0) q.push(v);
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}
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}
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}
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int main() {
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memset(h, -1, sizeof h);
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scanf("%d", &n);
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for (int a = 1; a <= n; a++) {
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int b;
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while (scanf("%d", &b), b) { // b==0时结束读入
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add(a, b);
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din[b]++; // 入度++
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}
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}
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// 拓扑序
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topsort();
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// 输出任意一组拓扑序
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for (int i = 0; i < n; i++) printf("%d ", path[i]);
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return 0;
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} |