UVA - 1395：Slim Span

Slim Span

来源：UVA
标签：图论
参考资料：《算法竞赛入门经典(第2版)》P358
相似题目：

题目

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.
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).

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), © 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.

输入

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 k-th 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).

输出

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.

输入样例

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

输出样例

1
20
0
-1
-1
1
0
1686
50

题目大意

给出一个n(n<=100)结点的图，求最大边减最小边的值尽量小的生成树。

解题思路

首先把边按权值从小到大排列。对于一个连续的边集区间[L, R]，如果这些边使得n个点全部连通，则一定存在一个苗条度不超过w[R]-w[L]的生成树，其中w[i]表示排序后第i条边的权值。从小到大枚举L，对于每个L，从小到大枚举R，同时用并查集将新进入[L, R]的边两端的点合并成一个集合，与Kruskal算法一样。当所有点连通时停止枚举R，换下一个L（并把R重置L），继续枚举。

参考代码

#include
#include
#define MAXN 5000
#define INF 0x3f3f3f3f
using namespace std;

int n,m;
struct Edge{
	int u,v,w;
}edge[MAXN];
int p[MAXN];

int cmp(Edge a, Edge b){
	return a.w<b.w;
}

int find(int x){
	return p[x]==x? x: p[x]=find(p[x]);
}

int main(){
	while(scanf("%d%d",&n,&m) && n){
		for(int i=0;i<m;i++){
			scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
		}
		sort(edge,edge+m,cmp);
		int L,R; 
		int ans=INF;
		for(L=0;L<m;L++){
			int dot_cnt=0;//计数已连通的点
			for(int i=0;i<=n;i++){//初始化n个节点的并查集 
				p[i]=i;
			}
			
			for(R=L;R<m;R++){
				int x=find(edge[R].u);
				int y=find(edge[R].v);
				if(x!=y){
					dot_cnt++;
					p[x]=y;
				}
				if(dot_cnt==n-1) break;//所有点都已连通 
			}
			
			if(dot_cnt==n-1){
				ans=min(ans,edge[R].w-edge[L].w);
			}else break;
		}
		if(ans==INF) printf("-1\n");
		else printf("%d\n",ans);
	}
	return 0;
}