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Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29087????Accepted Submission(s): 6698

Problem Description

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.?

Input

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.?

Output

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).?

Sample Input

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 0

-1 -1

Sample Output

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

Source

North Central North America 1997

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題意：

給n對點，每兩點之間有一條邊，判斷這些點是否可以構成一棵樹

思路一：

1.無環
2.除了根，所有的入度為 1，根的入度為 0
3.樹只有一個根，否則是森林

4.空樹也是樹

#include<bits/stdc++.h > using namespace std; int a,b; int f[100005]; int main() {int t=1;while(1){int edge=0;//樹的邊的個數 int node=0;//節點的個數 bool flag=0;//是否有回路 memset(f,0,sizeof(f));while(scanf("%d%d",&a,&b)&&a&&b){if(a<0||b<0)return 0;if(f[b]-1==1)//判斷是否有回路 flag=1;if(f[a]==0)node++;if(f[b]==0)node++;f[a]=1;f[b]=2;edge++;}//樹的節點個數等于邊的個數加一 if(flag==0&&node==edge+1)printf("Case %d is a tree.\n",t++); else printf("Case %d is not a tree.\n",t++); } }

思路二：并查集

?

#include <iostream> #include<cmath> #include<cstdio> #include<cstring> #define N 11000 using namespace std; int pre[N],vis[N],flag,maxn; void init() {for(int i=0;i<N;i++)pre[i]=i;memset(vis,0,sizeof(vis));maxn=0;flag=1; }int find(int x) {if(pre[x]==x) return x;return pre[x]=find(pre[x]); }void unions(int u,int v) {int t1=find(u);int t2=find(v);if(v!=t2||t1==t2) flag=0;if(t1!=t2) pre[t2]=t1; }int main() {int kase=0;int u,v;init();while(~scanf("%d%d",&u,&v)&&u>=0&&v>=0){if(u==0&&v==0){int tp=0;for(int i=1;i<=maxn;i++)//尋找根節點if(vis[i]&&i==pre[i]) tp++;if(tp<=1&&flag)printf("Case %d is a tree.\n",++kase);elseprintf("Case %d is not a tree.\n",++kase);init();continue;}maxn=max(maxn,max(u,v));vis[u]=vis[v]=1;unions(u,v);} }

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