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：
1 #include <cstdio> 2 #include <cmath> 3 #include <queue> 4 #include <cstring> 5 using namespace std; 6 const int INF = 0x3f3f3f3f; 7 const int MAXN = 1005; 8 int G[MAXN][MAXN], d[MAXN]; 9 bool in[MAXN]; 10 int n, p, k; 11 int spfa(int s, int t, int x) 12 { 13 memset(in, false, sizeof in); 14 memset(d, 0x3f, sizeof d); 15 queue<int> q; 16 d[s] = 0; 17 q.push(s); 18 in[s] = true; 19 while (!q.empty()) 20 { 21 int tmp = q.front(); q.pop(); 22 in[tmp] = false; 23 for (int i = 1; i <= n; i++) 24 { 25 int w = INF; 26 if (G[tmp][i] != INF) w = G[tmp][i] > x ? 1 : 0; 27 if (d[tmp] + w < d[i]) 28 { 29 d[i] = d[tmp] + w; 30 if (!in[i]) { in[i] = true; q.push(i); } 31 } 32 } 33 } 34 return d[t]; 35 } 36 bool check(int x) 37 { 38 int cnt = spfa(1, n, x); 39 return cnt <= k; 40 } 41 int main() 42 { 43 scanf("%d %d %d", &n, &p, &k); 44 int a, b, w, maxw = -INF; 45 memset(G, 0x3f, sizeof G); 46 for (int i = 0; i < p; i++) 47 { 48 scanf("%d %d %d", &a, &b, &w); 49 G[a][b] = G[b][a] = w; 50 maxw = max(maxw, w); 51 } 52 int l = 0, r = maxw, ans = INF; 53 while (l <= r) 54 { 55 int m = l + r >> 1; 56 if (check(m)) { ans = m; r = m - 1; } 57 else l = m + 1; 58 } 59 if (ans == INF) puts("-1"); 60 else printf("%d ", ans); 61 return 0; 62 }