POJ-3522 Slim Span(最小生成树)
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 8633 | Accepted: 4608 |
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
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=;
int f[];
int n,m;
struct Edge
{
int u,v,w;
};
Edge edge[]; bool cmp(Edge a,Edge b)
{
return a.w<b.w;
} int Find(int x)
{
int r = x;
while(r!=f[r])
{
r = f[r];
}
while(x!=f[x])
{
int j = f[x];
f[x] = f[r];
x = j;
}
return x;
} void merge2(int x,int y)
{
int fx=Find(x);
int fy=Find(y);
if(fx!=fy)
{
f[fy] = fx;
}
} int Cal(int x)
{
int i;
for(i=;i<=n;i++)
{
f[i] = i;
}
int mind=INF,maxd=-;
int cnt=;
for(i=x;i<m;i++)
{
int u=edge[i].u , v=edge[i].v , w=edge[i].w;
int fu=Find(u),fv=Find(v);
if(fu!=fv)
{
f[fu] = fv;
cnt++;
mind = min(mind,w);
maxd = max(maxd,w);
merge2(u,v);
}
if(cnt==n-)
break;
}
if(cnt == n-)
{
int ans = maxd-mind;
return ans;
}
return -;
} int main()
{
while(scanf("%d %d",&n,&m)!=EOF)
{
if(n==&&m==)
{
break;
}
int i,a,b,w;
for(i=;i<m;i++)
{
scanf("%d %d %d",&a,&b,&w);
edge[i].u=a;
edge[i].v=b;
edge[i].w=w;
}
sort(edge,edge+m,cmp);
int ans=INF;
for(i=;i<m;i++)
{
if(m-i<n-)
{
break;
}
int d = Cal(i);
if(d!=- && d<ans)
{
ans = d;
}
}
if(ans == INF)
printf("-1\n");
else
printf("%d\n",ans);
}
}
POJ-3522 Slim Span(最小生成树)的更多相关文章
- poj 3522 Slim Span (最小生成树kruskal)
http://poj.org/problem?id=3522 Slim Span Time Limit: 5000MS Memory Limit: 65536K Total Submissions ...
- POJ 3522 Slim Span 最小生成树,暴力 难度:0
kruskal思想,排序后暴力枚举从任意边开始能够组成的最小生成树 #include <cstdio> #include <algorithm> using namespace ...
- POJ 3522 Slim Span(极差最小生成树)
Slim Span Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 9546 Accepted: 5076 Descrip ...
- POJ 3522 ——Slim Span——————【最小生成树、最大边与最小边最小】
Slim Span Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 7102 Accepted: 3761 Descrip ...
- POJ 3522 Slim Span 最小差值生成树
Slim Span Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://poj.org/problem?id=3522 Description Gi ...
- POJ 3522 - Slim Span - [kruskal求MST]
题目链接:http://poj.org/problem?id=3522 Time Limit: 5000MS Memory Limit: 65536K Description Given an und ...
- POJ 3522 Slim Span
题目链接http://poj.org/problem?id=3522 kruskal+并查集,注意特殊情况比如1,0 .0,1.1,1 #include<cstdio> #include& ...
- POJ 3522 Slim Span 暴力枚举 + 并查集
http://poj.org/problem?id=3522 一开始做这个题的时候,以为复杂度最多是O(m)左右,然后一直不会.最后居然用了一个近似O(m^2)的62ms过了. 一开始想到排序,然后扫 ...
- POJ 3522 Slim Span (Kruskal枚举最小边)
题意: 求出最小生成树中最大边与最小边差距的最小值. 分析: 排序,枚举最小边, 用最小边构造最小生成树, 没法构造了就退出 #include <stdio.h> #include < ...
- uva1395 - Slim Span(最小生成树)
先判断是不是连通图,不是就输出-1. 否则,把边排序,从最小的边开始枚举最小生成树里的最短边,对每个最短边用Kruskal算法找出最大边. 或者也可以不先判断连通图,而是在枚举之后如果ans还是INF ...
随机推荐
- phpstrom识别phalcon框架模板文件的配置
- jquery判断对象是否显示或隐藏
if($('a.specail2').is(":visible")){ /**jquery判断对象是否显示或隐藏**/ $('a.one').hide(); $('a.specai ...
- Web前端总结(小伙伴的)
以下总结是我工作室的小伙伴的心得,可以参考一下 html+css知识点总结 HTMl+CSS知识点收集 1.letter-spacing和word-spacing的区别 letter-spacing: ...
- php通过curl扩展进行模拟登录(含验证码)
以下为本人工作中遇到的需要做的事情,之前也没怎么用过curl,查了好多资料,才稍微弄明白一点:本文所有内容只是自己平日工作的记录,仅供大家参考:<?php/*** 模拟登录*/header(&q ...
- 屏幕适配/autoLayout autoresizingMask
#pragma mark-- 屏幕适配/autoLayout autoresizingMask 1> 发展历程 代码计算frame -> autoreszing(父控件和子控件的关系) - ...
- Python 迭代器和列表解析
Python 迭代器和列表解析 1)迭代器 一种特殊的数据结构,以对象形式存在 >>> i1 = l1.__iter__() >>> i1 = iter(l1) 可 ...
- (转载)IQ 16.0 SP02起支持从压缩文件直接装载数据到表中
参考文档: http://m.blog.chinaunix.net/uid-16765068-id-4405877.htmlhttp://www.cnblogs.com/lichmama/p/4103 ...
- YARN框架详解
YARN框架详解 YARN官方解释 YARN是什么 The fundamental(定义) idea of YARN is to split(分开) up the functionalities(功能 ...
- Eclipse 修改 创建的Jsp的默认格式
Eclipse 的jsp模板修改 打开 eclipse 选择 Window -- Preferences
- shiro整合oauth
一.基本思路脑图 二.客户端shiro配置 shiro配置文件 <?xml version="1.0" encoding="UTF-8"?> < ...