POJ3259简单SPFA判负权回路

提交了好几次终于AC了，开始没仔细看题，后来才看见a bidirectional path between S and E that requires T seconds to traverse这句话！！原来路是双向的，虫洞是单向的，还有注意Two fields might be connected by more than one path. 这句话形同虚设，不用考虑这条边再次赋新值的取舍，直接再来一条新边就ok了。最近写图论比较多，不知道用vector好还是指针好= =还是请教一下老师比较好

#include <iostream>
#include <queue>
#include <vector>
using namespace std;
int n, m, p, w;
int b[100001], f[100001], e[100001];
int main()
{
    int i, x, y, z, t, tt;
    queue<int> que;
    vector<pair<int, int> > a[100001];
    vector<pair<int, int> > :: iterator it;
    scanf("%d", &w);
    while (w--)
    {
          scanf("%d%d%d", &n, &m, &p);
          memset(a, 0, sizeof(a));
          memset(b, 0, sizeof(b));
          memset(e, 0, sizeof(e));
          memset(f, 8, sizeof(f));
          for (i = 1; i <= m; i++)
          {
              cin >> x >> y >> z;
              a[x].push_back(make_pair(y, z));
              a[y].push_back(make_pair(x, z));
          }
          for (i = 1; i <= p; i++)
          {
              cin >> x >> y >> z;
              a[x].push_back(make_pair(y, -z));
          }
          que.push(1);
          b[1] = 1;
          f[1] = 0;
          while (!que.empty())
          {
                t = que.front();
                e[t]++;
                if (e[t] > n)
                {
                   b[0] = 1;
                   break;
                }
                for (it = a[t].begin(); it < a[t].end(); it++)
                {
                    i = it -> first;
                    if (f[i] > f[t] + it -> second)
                    {  
                       f[i] = f[t] + it -> second;
                       if (b[i] == 0)
                       {
                          b[i] = 1;
                          que.push(i);
                       }
                    }
                }
                b[t] = 0;
                que.pop();
          }      
          if (b[0] == 1)
             printf("YES\n");
          else
              printf("NO\n");
    }
    system("pause");
    return 0;
}