NYOJ 题目129树的判定(并查集)
来源:互联网 发布:投诉淘宝的电话是多少 编辑:程序博客网 时间:2024/06/04 19:37
树的判定
时间限制:1000 ms | 内存限制:65535 KB
难度:4
- 描述
A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following properties.
There is exactly one node, called the root, to which no directed edges point.
Every node except the root has exactly one edge pointing to it.
There is a unique sequence of directed edges from the root to each node.For example, consider the illustrations below, in which nodes are represented by circles and edges are represented by lines with arrowheads. The first two of these are trees, but the last is not.
In this problem you will be given several descriptions of collections of nodes connected by directed edges. For each of these you are to determine if the collection satisfies the definition of a tree or not.- 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. - 输出
- For each test case display the line "Case k is a tree." or the line "Case k is not a tree.", where k corresponds to the test case number (they are sequentially numbered starting with 1).
- 样例输入
6 8 5 3 5 2 6 4 5 6 0 08 1 7 3 6 2 8 9 7 5 7 4 7 8 7 6 0 03 8 6 8 6 4 5 3 5 6 5 2 0 0-1 -1
- 样例输出
Case 1 is a tree.Case 2 is a tree.Case 3 is not a tree.
- 来源
- POJ
- 上传者
- 张云聪
ac代码
#include<stdio.h>#include<string.h>int pre[10010],v[10010];int find(int x){int r=x;if(pre[x]==-1)return x;elsereturn find(pre[x]);}int main(){int a,b,c=0;while(scanf("%d%d",&a,&b)!=EOF,a!=-1||b!=-1){int k,flag,sum;if(!(a||b)){printf("Case %d is a tree.\n",++c);continue;}memset(v,0,sizeof(v));memset(pre,-1,sizeof(pre));k=flag=0;sum=1;if(a==b)flag=1;else{v[a]=v[b]=1;k+=2;}pre[b]=a;while(scanf("%d%d",&a,&b)!=EOF,a||b){int fx,fy;if(flag)continue;if(!v[a]){v[a]=1;k++;}if(!v[b]){v[b]=1;k++;}if(a==b){flag=1;continue;}fx=find(a);fy=find(b);if(fx==fy||fy!=b)//注意,fy!=b。b此时已经有父亲节点此时又有一个节点指向了b,也就是树中不存在入度大于一的节点 {flag=1;continue;}pre[fy]=fx;sum++;}if(flag||k!=sum+1){printf("Case %d is not a tree.\n",++c);}elseprintf("Case %d is a tree.\n",++c);}}
0 0
- NYOJ 题目129树的判定(并查集)
- NYOJ-129 树的判定(并查集)
- NYOJ 129 树的判定 (并查集)
- NYOJ-129 树的判定 并查集
- nyoj--129 树的判定(并查集)
- POJ 1308 Is It A Tree? && NYOJ 129 (树的判定+并查集)
- 树的判定(并查集)
- 树的判定(并查集)
- 树的判定 并查集
- NYOJ129 树的判定 【并查集】
- 树的判定----并查集
- 1896: 树的判定(并查集)
- nyoj 129树的判定
- nyoj 129树的判定
- hihoCoder 树结构判定(并查集)
- nyoj129树的判定,并查集(注意有向树的所有条件)
- 并查集的题目
- NYOJ--树的判定
- GNU Radio 模块扩展说明
- 【自考】数据结构之二叉树遍历
- CSS标签大全
- 打印shell脚本执行的命令到终端
- js实现列表无间隙循环滚动
- NYOJ 题目129树的判定(并查集)
- 【LeetCode】Algorithms 题集(四)
- 黑马程序员——面向对象(二)
- Jersey+Spring+Hibernate整合
- mysql-5.6.15-win32.zip 启动方式
- sles11之dns服务器配置
- 协议演变
- 常用颜色RGB值
- iOS iOS应用PUSH功能的实现