Is It A Tree? 并查集
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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.
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.
6 8 5 3 5 2 6 45 6 0 08 1 7 3 6 2 8 9 7 57 4 7 8 7 6 0 03 8 6 8 6 45 3 5 6 5 2 0 0-1 -1
Case 1 is a tree.Case 2 is a tree.Case 3 is not a tree.
首先这个问题吧,一看题目,这是一颗树吗?赤裸裸的告诉你是个并查集,再看输入,很明显,一父一子,说明就是普通的一个模板的调用,于是我们把模板愉快的调起来
#include <iostream>using namespace std;int par[3005],r[3005],num[3005];void init(){for(int i=0;i<3005;i++){par[i]=i;r[i]=0;num[i]=0;}}int find(int x){if(par[x]==x) return x;else return par[x]=find(par[x]);}void unite(int x,int y){x=find(x);y=find(y);if(x==y){return;}par[y]=x;}bool same(int x,int y){return find(x)==find(y);}int main(){int a,b,u=0;init();int flag=0;while(scanf("%d%d",&a,&b)!=EOF){if(a==-1&&b==-1){break;}if(flag==1&&a!=0&&b!=0){continue;}if(a==0&&b==0){int n=0;for(int j=1;j<3005;j++){if(num[j]&&find(j)==j){n++;}if(r[j]>1){flag=1;break;}}if(n>1){flag=1;}u++;if(flag==0){printf("Case %d is a tree.\n",u);}else{printf("Case %d is not a tree.\n",u);}flag=0;init();continue;//if(!unite(a,b)){//flag++;//}}if(a!=b&&find(a)==find(b)){flag=1;}else{num[a]=1;num[b]=1;r[b]++;unite(a,b);}}}
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