SPOJ TOPOSORT - Topological Sorting(拓扑排序)

Sandro is a well organised person. Every day he makes a list of things which need to be done and enumerates them from 1 to n. However, some things need to be done before others. In this task you have to find out whether Sandro can solve all his duties and if so, print the correct order.

Input

In the first line you are given an integer n and m (1<=n<=10000, 1<=m<=1000000). On the next m lines there are two distinct integers x and y, (1<=x,y<=10000) describing that job x needs to be done before job y.

Output

Print "Sandro fails." if Sandro cannot complete all his duties on the list. If there is a solution print the correct ordering, the jobs to be done separated by a whitespace. If there are multiple solutions print the one, whose first number is smallest, if there are still multiple solutions, print the one whose second number is smallest, and so on.

Sample Input & Output

Input
8 9
1 4
1 2
4 2
4 3
3 2
5 2
3 5
8 2
8 6

Ouptut
1 4 3 5 7 8 2 6 

Input
2 2
1 2
2 1

Output
Sandro fails.

---

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

size_t vertexNum, edgeNum;
vector<vector<int>> adj(10001);
vector<int> result, inDegree;
priority_queue<int, vector<int>, greater<int>> q;

ostream & operator<<(ostream & os, const vector<int> & v)
{
    for (size_t i = 0; i < v.size(); ++i){
        os << v[i];
        if (i != v.size() - 1) os << " ";
    }
    return os;
}

void topologicalSort()
{
    for (size_t i = 1; i <= vertexNum; ++i){
        if (inDegree[i] == 0)
            q.push(i);
    }

    while (!q.empty()) {
        int tmp = q.top();
        q.pop();
        result.push_back(tmp);
        for (auto i = adj[tmp].begin(); i != adj[tmp].end(); ++i){
            if (--inDegree[*i] == 0)
                q.push(*i);
        }
    }
    if (result.size() < vertexNum) cout << "Sandro fails." << endl;
    else cout << result << endl;
}


int main()
{
    cin >> vertexNum >> edgeNum;
    inDegree.assign(vertexNum + 1, 0);
    while (edgeNum--) {
        int point1, point2;
        cin >> point1 >> point2;
        adj[point1].push_back(point2);
        ++inDegree[point2];
    }
    topologicalSort();

    return 0;
}

这里应用了一个priority_queue<int, vector<int>, greater<int>>来保证每次按字典序最小的输出。