poj-3522 最小生成樹
Description
Given an undirected weighted graph G, you should find one of spanning trees specified as follows.
The graph G is an ordered pair (V, E), where V is a set of vertices {v1, v2, …, vn} and E is a set of undirected edges {e1, e2, …, em}. Each edge e ∈ E has its weight w(e).
A spanning tree T
Figure 5: A graph G and the weights of the edges
For example, a graph G
Figure 6: Examples of the spanning trees of G
There are several spanning trees for G. Four of them are depicted in Figure 6(a)~(d). The spanning tree Ta
Your job is to write a program that computes the smallest slimness.
Input
The input consists of multiple datasets, followed by a line containing two zeros separated by a space. Each dataset has the following format.
n | m | |
a1 | b1 | w1 |
? | ||
am | bm | wm |
Every input item in a dataset is a non-negative integer. Items in a line are separated by a space. n is the number of the vertices and m the number of the edges. You can assume 2 ≤ n ≤ 100 and 0 ≤ m ≤ n(n ? 1)/2. ak and bk (k = 1, …,m) are positive integers less than or equal to n, which represent the two vertices vak and vbk connected by the kth edge ek. wk is a positive integer less than or equal to 10000, which indicates the weight of ek. You can assume that the graph G = (V, E) is simple, that is, there are no self-loops (that connect the same vertex) nor parallel edges (that are two or more edges whose both ends are the same two vertices).
Output
For each dataset, if the graph has spanning trees, the smallest slimness among them should be printed. Otherwise, ?1 should be printed. An output should not contain extra characters.
Sample Input
4 5 1 2 3 1 3 5 1 4 6 2 4 6 3 4 7 4 6 1 2 10 1 3 100 1 4 90 2 3 20 2 4 80 3 4 40 2 1 1 2 1 3 0 3 1 1 2 1 3 3 1 2 2 2 3 5 1 3 6 5 10 1 2 110 1 3 120 1 4 130 1 5 120 2 3 110 2 4 120 2 5 130 3 4 120 3 5 110 4 5 120 5 10 1 2 9384 1 3 887 1 4 2778 1 5 6916 2 3 7794 2 4 8336 2 5 5387 3 4 493 3 5 6650 4 5 1422 5 8 1 2 1 2 3 100 3 4 100 4 5 100 1 5 50 2 5 50 3 5 50 4 1 150 0 0
Sample Output
1 20 0 -1 -1 1 0 1686 50
kruskal 求最小生成樹 暴力枚舉
1 #include <cstdio> 2 #include <cstring> 3 #include <string> 4 #include <algorithm> 5 #include <queue> 6 using namespace std; 7 8 const int maxn = 5e4 + 10; 9 const int INF = 0x7fffffff; 10 int fa[510], vis[maxn]; 11 int n, m; 12 struct node { 13 int u, v, w; 14 } qu[maxn]; 15 int cmp(node a, node b) { 16 return a.w < b.w; 17 } 18 int Find(int x) { 19 return fa[x] == x ? x : fa[x] = Find(fa[x]); 20 } 21 int combine(int x, int y) { 22 int nx = Find(x); 23 int ny = Find(y); 24 if(nx != ny) { 25 fa[nx] = ny ; 26 return 1; 27 } 28 return 0; 29 } 30 int kruskal(int x) { 31 int big = -INF, small = INF, k = 0 ; 32 for (int i = x ; i < m ; i++) { 33 if (combine(qu[i].u, qu[i].v)) { 34 k++; 35 big = max(big, qu[i].w); 36 small = min(small, qu[i].w); 37 } 38 } 39 if (k!=n-1) return INF; 40 if (k==1) return 0; 41 return big - small; 42 } 43 int main() { 44 while(scanf("%d%d", &n, &m) != EOF) { 45 if (n == 0 && m == 0) break; 46 for (int i = 0 ; i < m ; i++) 47 scanf("%d%d%d", &qu[i].u, &qu[i].v, &qu[i].w); 48 sort(qu, qu + m, cmp); 49 int ans = INF; 50 for (int i = 0 ; i < m; i++) { 51 for (int j = 0 ; j <= n ; j++) fa[j] = j; 52 ans = min(ans, kruskal(i)); 53 } 54 if (ans==INF) printf("-1\n"); 55 else printf("%d\n", ans); 56 } 57 return 0; 58 }
poj-3522 最小生成樹