HDOJ 1325 Is It A Tree?

Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22086    Accepted Submission(s): 4996

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 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

 

Sample Output

Case 1 is a tree.Case 2 is a tree.Case 3 is not a tree.
因为出在并查集的专题，所以我执念要用并查集写，然后这道题WA了45次。真的是45次。到现在我也不知道哪错了。最后没办法，只能请教巨巨，最后不用并查集既简单又不容易错。怀疑人生- -
#include<iostream>#include<cstdio>#include<cstring>using namespace std;int f[100005];int main(){    int a,b,flag,i,j;    int t=1;    while(1)    {        j=0;        i=0;        flag=0;        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)//判断是否有回路                flag=1;            if(f[a]==0)                j++;            if(f[b]==0)                j++;            f[a]=1;f[b]=2;i++;        }        if(flag==0&&j==i+1)//n个点组成一个数，则点的个数=边的个数加一            printf("Case %d is a tree.\n",t++);            else printf("Case %d is not a tree.\n",t++);    }}

下面是WA代码，过段时间我再回头看
#include<stdio.h>int fa[100010],flag[100010],FLAG;void Init(){for(int i=0;i<100000;i++){fa[i]=i;flag[i]=0;}}int find(int x){if(x!=fa[x])fa[x]=find(fa[x]);return fa[x];}int mix(int x,int y){int p=find(x);int q=find(y);if(p==q||q!=y)FLAG=1;elsefa[q]=p;}int main(){int cnt,t=0;int a,b;while(~scanf("%d%d",&a,&b)){FLAG=0;cnt=0;if(a==-1&&b==-1)break;if(a==0&&b==0){printf("Case %d is a tree.\n",++t);continue;}Init();if(find(a)==find(b))FLAG=1;fa[find(b)]=find(a); int j,k;while(scanf("%d%d",&j,&k)&&j||k){mix(j,k);}for(int i=1;i<100000;i++)if(fa[i]!=i)flag[find(i)]++;for(int i=1;i<100000;i++)if(flag[i])cnt++;if(cnt!=1)FLAG=1;if(FLAG==1)printf("Case %d is not a tree.\n",++t);else printf("Case %d is a tree.\n",++t);}return 0;}