POJ 3522 Slim Span【最大边权差最小生成树Kruskal】

http://poj.org/problem?id=3522
Slim Span
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 9228 Accepted: 4905
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 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 Ta 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 Ta 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.

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
Source

Japan 2007

题意：最大边权差最小生成树的意思是，一个生成树中的最大边和最小边之差的值是所有生成树中最小的。

分析：据说可以用扫描线做，打算学习一下。其实题目中的要求和kruskal的做法如出一辙，因为kruskal优先选择就是权值最小可以构建的边。所以我们可以根据权值区间的起始位置到末尾位置进行枚举，来进行多个kruskal的操作，取答案最小的情况。当然某一状态的生成树建造不出来的话，那么这一状态当然是不计入答案的。

#include
#include
#include
#include
#include
using namespace std;
const int maxn = 1e4 + 10;
#define INF 0x7fffffff
struct edge { int u, v, w; }e[maxn];
int f[105];
int find(int i) {
    return f[i] == i ? i : f[i] = find(f[i]);
}
//bool operator<(const edge &a, const edge &b) {return a.w < b.w;}
int cmp(edge a, edge b) { return a. w < b.w; }
int main()
{
    int n, m;
    while (~scanf("%d%d", &n, &m) && (n + m)) {
        for (int i = 0; i < m; i++) scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w);
        //sort(e, e + m);
        sort(e, e + m, cmp); int ans = INF;
        for (int i = 0; i < m; i++) {
            for (int j = 1; j <= n; j++) {
                f[j] = j;
            }
            int res = n;
            for (int j = i; j < m; j++) {
                if (find(e[j].u) != find(e[j].v)) {
                    res--; f[find(e[j].u)] = find(e[j].v);
                    if (res == 1) {
                        ans = min(ans, e[j].w - e[i].w);
                        //cout << 1 << endl;
                        break;
                    }
                }
            }
        }
        printf("%d\n", ans == INF ? -1 : ans);
    }
    return 0;
}