POJ 2367 Genealogical tree 拓扑排序

本题属于最基础的拓扑排序。大意就是，给出一个数n， 然后底下n行，编号1到n, 每行输入几个数，保证结果中该行的编号要在这几个数前面。

/*
ID: sdj22251
PROG: calfflac
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAX 100000000
#define LOCA
#define PI acos(-1.0)
using namespace std;
int n, m;
int in[201];
int ans[201];
vector<int>v[201];
void topsort()
{
    int i, ct = 0;
    queue<int>q;
    for(i = 1; i <= n; i++)
    {
        if(in[i] == 0)
        q.push(i);
    }
    while(!q.empty())
    {
        int first = q.front();
        ans[ct++] = first;
        q.pop();
        int size = v[first].size();
        for(i = 0; i < size; i++)
        {
            int can_reach = v[first][i];
            in[can_reach]--;
            if(in[can_reach] == 0)
            q.push(can_reach);
        }
    }
}
int main()
{
#ifdef LOCAL
    freopen("ride.in","r",stdin);
    freopen("ride.out","w",stdout);
#endif
    int i, tmp;
    scanf("%d", &n);
    for(i = 1; i <= n; i++)
    {
        while(1)
        {
            scanf("%d", &tmp);
            if(tmp == 0)
            break;
            v[i].push_back(tmp);
            in[tmp]++;
        }
    }
    topsort();
    for(i = 0; i < n - 1; i++)
    printf("%d ", ans[i]);
    printf("%d\n", ans[n - 1]);
    return 0;
}