Slim Span
Time Limit: 5000MSMemory Limit: 65536K
Total Submissions: 6685Accepted: 3544
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 weightw(e).
A spanning tree T is a tree (a connected subgraph without cycles) which connects all the n vertices withn 1 edges. The slimness of a spanning tree T is defined as the difference between the largest weight and the smallest weight among then 1 edges of T.
Figure 5: A graph G and the weights of the edges
For example, a graph G in Figure 5(a) has four vertices {v1,v2, v3, v4} and five undirected edges {e1,e2, e3, e4, e5}. The weights of the edges arew(e1) = 3, w(e2) = 5, w(e3) = 6, w(e4) = 6, w(e5) = 7 as shown in Figure 5(b).
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 treeTa in Figure 6(a) has three edges whose weights are 3, 6 and 7. The largest weight is 7 and the smallest weight is 3 so that the slimness of the treeTa is 4. The slimnesses of spanning trees Tb,Tc and Td shown in Figure 6(b), (c) and (d) are 3, 2 and 1, respectively. You can easily see the slimness of any other spanning tree is greater than or equal to 1, thus the spanning tree Td in Figure 6(d) is one of the slimmest spanning trees whose slimness is 1.
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.
nm
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 andbk (k = 1, …, m) are positive integers less than or equal ton, which represent the two vertices vak andvbk connected by the kth edge ek.wk is a positive integer less than or equal to 10000, which indicates the weight ofek. 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 51 2 31 3 51 4 62 4 63 4 74 61 2 101 3 1001 4 902 3 202 4 803 4 402 11 2 13 03 11 2 13 31 2 22 3 51 3 65 101 2 1101 3 1201 4 1301 5 1202 3 1102 4 1202 5 1303 4 1203 5 1104 5 1205 101 2 93841 3 8871 4 27781 5 69162 3 77942 4 83362 5 53873 4 4933 5 66504 5 14225 81 2 12 3 1003 4 1004 5 1001 5 502 5 503 5 504 1 1500 0
Sample Output
1200-1-110168650
Source
题目链接：?id=3522题目大意：n个点，m条边，求一个生成树，使得其最大边权与最小边权的差最小，输出这个差值，，如果不能构成生成树，输出－1题目分析：用Kruskal的贪心思想加枚举，从第一条边开始，每次去掉一条边求生成树更新差值的最小值，详细见程序#include <cstdio>#include <algorithm>#include <cstring>using namespace std;int const INF = 0xffffff;int const MAX = 105;int n, m, ans;int fa[MAX], rank[MAX];bool flag, ok;struct Edge{int u, v, w;}e[6000];bool cmp(Edge a, Edge b){return a.w < b.w;}void UFset(){for(int i = 0; i <= n; i++)fa[i] = i;memset(rank, 0, sizeof(rank));}int Find(int x){return x == fa[x] ? x : fa[x] = Find(fa[x]);}void Union(int a, int b){int r1 = Find(a);int r2 = Find(b);if(r1 == r2)return;if(rank[r1] > rank[r2])fa[r2] = r1;else{fa[r1] = r2;if(rank[r1] == rank[r2])rank[r1]++;}}int Kruskal(int v0){int num = 0, mi = INF, ma = -1;UFset();for(int i = v0; i < m; i++){int u = e[i].u;int v = e[i].v;if(Find(u) != Find(v)){mi = min(mi, e[i].w); //记录最大边权ma = max(ma, e[i].w); //记录最小边权Union(u, v);num++;}if(num == n – 1) //够造成一颗生成树{ok = true; //ok表示存在一颗树flag = true; //flag表示当前情况下存在一棵树break;}}return ma – mi;}int main(){while(scanf("%d %d", &n, &m) != EOF && (m + n)){ok = false;for(int i = 0; i < m; i++)scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w);sort(e, e + m, cmp);int x, y, ans = INF;for(int i = 0; i < m; i++) //枚举{flag = false;int tmp = Kruskal(i);if(flag)ans = min(ans, tmp);}printf("%d\n", !ok ? -1 : ans);}}
那段岁月，无论从何种角度读你，你都完美无缺，
