POJ 3662 Telephone Lines(二分+最短路)

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最小化第K大值。

让我怀疑人生的一题目，我有这么笨？

 1 #include <cstdio>
 2 #include <queue>
 3 #include <cstring>
 4 #include <vector>
 5 #include <functional>
 6 using namespace std;
 7 #define maxv 1010
 8 #define maxl 1000000
 9 struct edge
10 {
11     int to, cost;
12     edge(){}
13     edge(int to, int cost) : to(to), cost(cost){}
14 };
15 typedef pair<int, int> P;
16 vector<edge> G[maxv];
17 int d[maxv];
18 int V, E;
19 int dij(int s, int x) {
20     priority_queue<P, vector<P>, greater<P> > que;
21     memset(d, 0X3f, V*sizeof(int));
22     d[s] = 0;
23     que.push(P(0, s));
24     while (!que.empty()) {
25        P p = que.top(); que.pop();
26         int v = p.second;
27         if (d[v] < p.first) continue;
28         for (int i = 0; i < G[v].size(); ++i)
29         {
30             edge e = G[v][i];
31             int new_d = d[v] + (e.cost >= x ? 1 : 0);
32             if (d[e.to] > new_d)
33             {
34                 d[e.to] = new_d;
35                 que.push(P(d[e.to], e.to));
36             }
37         }
38     }
39     return d[V-1];
40 }
41 int main(void) {
42     freopen("in.txt", "r", stdin);
43     freopen("out.txt", "w", stdout);
44     int K;
45     scanf("%d%d%d", &V, &E, &K);
46     for (int i = 0; i < E; ++i) {
47         int A, B, L;
48         scanf("%d%d%d", &A, &B, &L);
49         --A, --B;
50         G[A].push_back(edge(B,L));
51         G[B].push_back(edge(A,L));
52     }
53     int l = 0, u = maxl+9;
54     for (;(u-l) > 1;) {
55         int m = (u+l) >> 1;
56         if (dij(0, m) > K) {
57             l = m;
58         } else {
59             u = m;
60         }
61     }
62     printf("%d\n", (l>maxl ? -1 : l));
63     return 0;
64 }

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