POJ 2395 Out of Hay

http://poj.org/problem?id=2395
裸最小生成树输出树中最大cost的边值
直接prim
 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <queue>
 5 #include <algorithm>
 6 #define READ() freopen("in.txt", "r", stdin);
 7 #define MAXV 2007
 8 #define MAXE 20007
 9 #define INF 0x3f3f3f3f3f3f3f3f
10 
11 using namespace std;
12 int N, M;
13 typedef long long ll;
14 typedef pair<ll, int> P;
15 
16 struct Edge
17 {
18     int to, next;
19     ll cost;
20     Edge() {}
21     Edge(int to, ll cost, int next) : to(to), cost(cost), next(next) {}
22 }edge[MAXE];
23 
24 int head[MAXV];
25 int num = 0;
26 void Add(int from, int to, ll cost)
27 {
28     edge[num] = Edge(to, cost, head[from]);
29     head[from] = num++;
30 }
31 
32 int prim()
33 {
34     ll dist[MAXV];
35     ll rec[MAXV], r = 0;
36     bool use[MAXV];
37     priority_queue<P, vector<P>, greater<P> > que; // 从小到大排啊
38     fill(dist, dist+MAXV, INF);
39     fill(use, use+MAXV, false);
40     dist[1] = 0;
41     que.push(P(dist[1], 1));
42     while (!que.empty())
43     {
44         P p = que.top();
45         que.pop();
46         ll d = p.first;
47         int v = p.second;
48         if (!use[v]) rec[r++] = dist[v];
49         use[v] = true;
50         if (dist[v] < d) continue;
51         int t = head[v];
52         while (t != -1)
53         {
54             Edge e = edge[t];
55             if (!use[e.to] && dist[e.to] > e.cost)
56             {
57                 dist[e.to] = e.cost;
58                 que.push(P(dist[e.to], e.to));
59             }
60             t = e.next;
61         }
62     }
63     sort(rec, rec+r);
64     return rec[r-1];
65 }
66 
67 
68 
69 int main()
70 {
71     READ()
72     scanf("%d%d", &N, &M);
73     memset(edge, -1, sizeof(edge));
74     memset(head, -1, sizeof(head));
75     for (int i = 0; i < M; i++)
76     {
77         int from, to;
78         ll cost;
79         scanf("%d%d", &from, &to);
80         scanf("%lld", &cost);
81         Add(from, to, cost);
82         Add(to, from, cost);
83     }
84     ll ans = prim();
85     cout << ans << endl;
86 }