zoukankan      html  css  js  c++  java
  • 树的判定

    判断输入的是不是一棵树

    输入

    The input will consist of a sequence of descriptions (test cases) followed by a pair of negative integers. Each test case will consist of a sequence of edge descriptions followed by a pair of zeroes Each edge description will consist of a pair of integers; the first integer identifies the node from which the edge begins, and the second integer identifies the node to which the edge is directed. Node numbers will always be greater than zero.

    The number of test cases will not more than 20,and the number of the node will not exceed 10000.
    The inputs will be ended by a pair of -1.样例输入:6 8  5 3  5 2  6 4 5 6  0 0

    8 1  7 3  6 2  8 9  7 5 7 4  7 8  7 6  0 0

    3 8  6 8  6 4 5 3  5 6  5 2  0 00 03 8 6 8 0 0
    -1 -1

    样例输出:

    Case 1 is a tree.
    Case 2 is a tree.
    Case 3 is not a tree.

    Case 4 is a tree.

    Case 5 is not a tree.

    代码如下:

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 using namespace std;
     5 const int N=10000+100;
     6 int in[N],vis[N];
     7 int p[N];
     8 int find(int x) {return x==p[x]? x: p[x]=find(p[x]);}
     9 int main()
    10 {
    11     int a,b,n=0,cs=1,i;
    12     bool istree=1;
    13     for(i=0;i<N;i++) p[i]=i;
    14     memset(vis,0,sizeof(vis));
    15     while(cin>>a>>b)
    16     {
    17         if(a==b&&a==-1) break;
    18         if(a!=0&&b!=0)
    19         {
    20             n=max(n,max(a,b));
    21             in[b]++;vis[a]=1;vis[b]=1;
    22             if(istree&&in[b]>1)
    23             {
    24                 istree=0;printf("Case %d is not a tree.\n",cs++);
    25             }
    26             if(istree)
    27             {
    28                 int x=find(a),y=find(b);
    29                 if(x!=y)
    30                 p[y]=x;
    31             }
    32         }
    33         else
    34         {
    35             if(istree)
    36             {
    37                 int t=0,s=0;
    38                 for(i=0;i<=n;i++)
    39                 {
    40                     if(vis[i]&&i==find(i)) t++;
    41                     if(t>1) break;
    42                     if(vis[i]&&in[i]==0) s++;
    43                     if(s>1) break;
    44                 }
    45                 if(i==n+1&&s==1)
    46                 printf("Case %d is a tree.\n",cs++);
    47                 else
    48                 printf("Case %d is not a tree.\n",cs++);
    49 
    50             }
    51             istree=1;
    52             for(i=0;i<n;i++) p[i]=i;
    53             memset(vis,0,sizeof(vis));
    54             memset(in,0,sizeof(in));
    55         }
    56     }
    57     return 0;
    58 }
  • 相关阅读:
    vue-cli的使用
    修饰模式(Decorator结构型)C#简单例子
    c#继承中的函数调用
    c#桥接模式(bridge结构模式)
    c#浅谈反射内存的处理
    C#中的try catch finally
    C#微信公众号开发系列教程(接收事件推送与消息排重)
    用 C# 读取二进制文件
    c#语言-多线程中的锁系统(一)
    .NET程序内,访问私有或者保护成员的技巧
  • 原文地址:https://www.cnblogs.com/qijinbiao/p/2215448.html
Copyright © 2011-2022 走看看