hdu 1325 Is It A Tree?(并查集)
2016-10-07 23:14
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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 Input6 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 OutputCase 1 is a tree. Case 2 is a tree. Case 3 is not a tree. 大致题意: 给出几个关系,问能不能构成树。 不能成环,不能是森林。一个点只能有一个父亲。 思路分析: 新加入的两个点不能有相同的祖先。这个是防止成环。 所有的点必须有相同的根节点。 ac代码: #include<bits/stdc++.h> using namespace std; int parent[50005]; int vis[50005]; int ans; int n,m,k; int GetParent(int a) { if(a!=parent[a]) parent[a]=GetParent(parent[a]); return parent[a]; } void Merge(int a,int b) { int p1=GetParent(a); int p2=GetParent(b); if(p1==p2) { ans=0; return ; } if(p2!=b) //防止有多个父亲。 { ans=0; return ; } parent[p2]=p1; } int main() { int minn,maxx; int top=0; int i,j; while(true) { ans=1; cin>>n>>m; if(n<0 && m<0) break; minn=min(n,m); maxx=max(n,m); cout<<"Case "<<++top<<" "; for(i=0;i<=1002;i++) { parent[i]=i; vis[i]=0; } Merge(n,m); vis =vis[m]=1; while(true) { cin>>n>>m; if(n==0 && m==0) break; if(n<minn) minn=n; if(m<minn) minn=m; if(n>maxx) maxx=n; if(m>maxx) maxx=m; Merge(n,m); vis =vis[m]=1; } int flag=0; for(i=minn;i<=maxx;i++) //保证只有一个根节点。(不是森林~) { if(GetParent(i)==i && vis[i]) flag++; if(flag>1) { ans=0; break; } } if(ans) cout<<"is a tree."<<endl; else cout<<"is not a tree."<<endl; } } |
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