poj3662 Telephone Lines

思路：

二分+最短路。最短路也可以用来计算从a到达b所需的边权不超过x的边的数量。

实现：

 #include <cstdio>
 #include <cmath>
 #include <queue>
 #include <cstring>
 using namespace std;
 const int INF = 0x3f3f3f3f;
 const int MAXN = ;
 int G[MAXN][MAXN], d[MAXN];
 bool in[MAXN];
 int n, p, k;
 int spfa(int s, int t, int x)
 {
     memset(in, false, sizeof in);
     memset(d, 0x3f, sizeof d);
     queue<int> q;
     d[s] = ;
     q.push(s);
     in[s] = true;
     while (!q.empty())
     {
         int tmp = q.front(); q.pop();
         in[tmp] = false;
         for (int i = ; i <= n; i++)
         {
             int w = INF;
             if (G[tmp][i] != INF) w = G[tmp][i] > x ?  : ;
             if (d[tmp] + w < d[i])
             {
                 d[i] = d[tmp] + w;
                 if (!in[i]) { in[i] = true; q.push(i); }
             }
         }
     }
     return d[t];
 }
 bool check(int x)
 {
     int cnt = spfa(, n, x);
     return cnt <= k;
 }
 int main()
 {
     scanf("%d %d %d", &n, &p, &k);
     int a, b, w, maxw = -INF;
     memset(G, 0x3f, sizeof G);
     for (int i = ; i < p; i++)
     {
         scanf("%d %d %d", &a, &b, &w);
         G[a][b] = G[b][a] = w;
         maxw = max(maxw, w);
     }
     int l = , r = maxw, ans = INF;
     while (l <= r)
     {
         int m = l + r >> ;
         if (check(m)) { ans = m; r = m - ; }
         else l = m + ;
     }
     if (ans == INF) puts("-1");
     else printf("%d\n", ans);
     return ;
 }