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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 }
POJ 2395 Out of Hay
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