zoukankan      html  css  js  c++  java
  • 【UVA 1395】 Slim Span (苗条树)

    【题意】

      求一颗生成树,满足最大边和最小边之差最小

    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 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).
    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

    【分析】

      做完这道题你就变苗条了的意思。。

      [沉迷打机,日渐消瘦。

      正题->_->先把边排序,枚举最小边,然后就是让最长边最短了咯,就是最小瓶颈生成树,kruskal做最小生成树就好了。

      复杂度:m^2

     1 #include<cstdio>
     2 #include<cstdlib>
     3 #include<cstring>
     4 #include<iostream>
     5 #include<algorithm>
     6 #include<queue>
     7 using namespace std;
     8 #define INF 0xfffffff
     9 #define Maxn 110
    10 
    11 struct node
    12 {
    13     int x,y,c;
    14 }t[Maxn*Maxn];
    15 int n,m;
    16 
    17 bool cmp(node x,node y) {return x.c<y.c;}
    18 int mymax(int x,int y) {return x>y?x:y;}
    19 int mymin(int x,int y) {return x<y?x:y;}
    20 
    21 int fa[Maxn];
    22 int ffa(int x)
    23 {
    24     if(fa[x]!=x) fa[x]=ffa(fa[x]);
    25     return fa[x];
    26 }
    27 
    28 int main()
    29 {
    30     while(1)
    31     {
    32         scanf("%d%d",&n,&m);
    33         if(n==0&&m==0) break;
    34         for(int i=1;i<=m;i++) scanf("%d%d%d",&t[i].x,&t[i].y,&t[i].c);
    35         sort(t+1,t+1+m,cmp);
    36         int cnt=0,ans=INF;
    37         for(int i=1;i<=m;i++)
    38         {
    39             int cnt=0;
    40             for(int j=1;j<=n;j++) fa[j]=j;
    41             for(int j=i;j<=m;j++)
    42             {
    43                 if(ffa(t[j].x)!=ffa(t[j].y))
    44                 {
    45                     fa[ffa(t[j].x)]=ffa(t[j].y);
    46                     cnt++;
    47                 }
    48                 if(cnt==n-1) {ans=mymin(ans,t[j].c-t[i].c);break;}
    49             }
    50         }
    51         if(ans==INF) printf("-1
    ");
    52         else printf("%d
    ",ans);
    53     }
    54     return 0; 
    55 }
    View Code

    跟LA3887不是一样的么,LA有毒啊狂wa。。蓝书的陈锋的代码也是wa的啊smg!!

    2016-11-01 21:24:53

  • 相关阅读:
    Android之上下文context
    如果简单的记录,就可以为这个世界创造更多的财富,那么还有什么理由不去写博客呢? — 读<<黑客与画家>> 有感
    【最新最全】为 iOS 和 Android 的真机和模拟器编译 Luajit 库
    【Graphql实践】使用 Apollo(iOS) 访问 Github 的 Graphql API
    【最新】LuaJIT 32/64 位字节码,从编译到使用全纪录
    简陋的swift carthage copy-frameworks 辅助脚本
    【自问自答】关于 Swift 的几个疑问
    【读书笔记】The Swift Programming Language (Swift 4.0.3)
    【读书笔记】A Swift Tour
    【趣味连载】攻城狮上传视频与普通人上传视频:(一)生成结构化数据
  • 原文地址:https://www.cnblogs.com/Konjakmoyu/p/6021059.html
Copyright © 2011-2022 走看看