Sicily 1504. Slim Span

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1504. Slim Span

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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, , v_n} and E is a set of undirected edges {e₁, e₂, , e_m}. Each edgee ? E has its weight w(e).

A spanning tree T is a tree (a connected subgraph without cycles) which connects all the n vertices with n ? 1 edges. The slimness of a spanning tree T is defined as the difference between the largest weight and the smallest weight among the n ? 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 are w(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 tree T_a 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 tree T_a is 4. The slimnesses of spanning trees T_b, T_c and T_d 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 a₁b₁w₁ ? a_mb_mw_m

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. a_k and b_k (k = 1, , m) are positive integers less than or equal ton, which represent the two vertices v_{a_k} and v_{b_k} connected by the kth edge e_k. w_k is a positive integer less than or equal to 10000, which indicates the weight of e_k. 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

de style="color: rgb(51, 51, 51); font-size: 18.01801872253418px; line-height: 27.027027130126953px;" >

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

de>

Sample Output

de style="color: rgb(51, 51, 51); font-size: 18.01801872253418px; line-height: 27.027027130126953px;" >

1200-1-110168650

de>

Problem Source

Tokyo 2007

求最大边和最小边之差最小的生成树，答案输出最小差。

取定某一条边，以此边开始用kruskal方法做最小生成树，那么得到的最大边是尽量小的最大边。

于是，将所有的边从小到大排序，并遍历各边，以各边开始（此时最小边已经确定）做最小生成树，得到对应的尽量小的最大边，计算差值，更新答案。

#include <algorithm>#include <iostream>#include <string>#include <stdio.h>#include <queue>#include <string.h>#include <vector>#include <iomanip>#include <map>#include <stack>#include <functional>#include <list>#include <math.h>using namespace std;#define MAX_N 105#define INF 2000000000struct edge{    int u;    int v;    int cost;};vector<edge> G;double x;int n;int m;int par[MAX_N];void init(int n) {    for (int i = 1; i <= n; i++) {        par[i] = i;    }}int find(int x) {    if (par[x] == x) return x;    par[x] = find(par[x]);    return par[x];}void unite(int x, int y) {    x = find(x);    y = find(y);        if (x == y) return;    par[x] = y;}bool same(int x, int y) {    return find(x) == find(y);}bool cmp(edge e1, edge e2) {    return e1.cost < e2.cost;}int kruskal(int st) {    init(n);        edge lastEdge;    int eNum = 0;    for (int i = st; i < m; i++) {        edge e = G[i];        if (!same(e.u, e.v)) {            unite(e.u, e.v);            eNum++;            if (eNum == n - 1) {                return G[i].cost - G[st].cost;                break;            }        }    }    return INF;}int main() {    std::ios::sync_with_stdio(false);    while (1) {        cin >> n >> m;        if (n == 0 && m == 0) break;        G.clear();        G.resize(m);        for (int i = 0; i < m; i++)  cin >> G[i].u >> G[i].v >> G[i].cost;        int ans = INF;        sort(G.begin(), G.end(), cmp);        for (int i = 0; i <= m - (n - 1); i++) {            int newAns = kruskal(i);            if (newAns < ans) ans = newAns;        }        cout << ans << endl;    }    //getchar();    //getchar();        return 0;}

此处贴上自己写的错误代码：

int kruskal(int st) {    init(n);        edge lastEdge;    int eNum = 0;    for (int i = st; i < m; i++) {        edge e = G[i];        if (!same(e.u, e.v)) {            unite(e.u, e.v);            eNum++;            lastEdge = e;            if (eNum == n - 1) break;        }    }    return lastEdge.cost - G[st].cost;}

不同之处就在于上面的函数，注意到在错误代码中，如果st达到了一定大的值，以st到m-1这些边有可能是不能生成最小生成树的（边会越来越少），但错误代码中在这种情况下依然会返回差值，这就是错误原因。

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