poj3522Slim Span(最小生成樹性質)

2019-11-14 11:22:18

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題目鏈接

Slim Span

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 7870		Accepted: 4191

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 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 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.

n	m
a₁	b₁	w₁
	?
a_m	b_m	w_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 to n, 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

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

Japan 2007

題意：

給出一個圖，若圖連通，則求最大邊與最小邊差值最小的生成樹，輸出最小差值。否則輸出-1.

題解：

最小生成樹有一個很重要的性質：在構造生成樹時有可能選擇不同的邊，但最小生成樹的權是唯一的！所以在用kruskal算法時第一次加入的必然是最小生成樹的最小邊權值，最小邊確定后，最小生成樹的最大邊的權值是所以生成樹中最小的，于是只要枚舉最小邊，然后求最小生成樹，就可以得到最大邊，只要每次更新最優解就行了。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<vector>#include<queue>#include<stack>using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define pb push_back#define fi first#define se secondtypedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;const int inf=0x3fffffff;const ll mod=1000000007;const int maxn=100+10;int pa[maxn];struct edge{    int u,v,w;    bool Operator <(const edge&b)const{        return w<b.w;    }}e[maxn*maxn];int find(int x){    return pa[x]==x? x:pa[x]=find(pa[x]);}int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        if(n==0&&m==0) break;        int ans=inf;        for(int i=0;i<m;i++)        {            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            e[i].u=u,e[i].v=v,e[i].w=w;        }        sort(e,e+m);        rep(i,0,m)        {            rep(k,1,n+1) pa[k]=k;            int mi=e[i].w;            pa[e[i].v]=e[i].u;            int cnt=1;            rep(j,i+1,m)            {                int fu=find(e[j].u),fv=find(e[j].v);                if(fu==fv)                    continue;                pa[fu]=fv;                cnt++;                if(cnt==n-1)                {                    ans=min(ans,e[j].w-mi);                    break;                }            }            if(i==0&&cnt<n-1)            {                ans=-1;                break;            }            if(cnt<n-1) break;        }        if(!m) ans=-1;        if(m==1)        {            if(n==2) ans=0;            else ans=-1;        }        cout << ans << endl;    }    return 0;}

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