HDU 1325 poj 1308 is it a tree? (Query set)

This question is the bare parallel query set. The key lies in the judgment of several types of trees.

1. Empty tree is a tree 2. Forest is not a tree 3. No ring

Or from the perspective of the inbound degree: 1, no ring; 2, except for the root, all the inbound degrees are 1, root inbound degrees are 0; 3, this structure has only one root, otherwise it will be the forest.

This question was originally made for poj 1308 during the summer vacation, but not for HDU. The empty tree is not considered.

Use the check set to determine the number of forests. Note that the numbers are random, not in order ....

/*input:0 01 1 0 01 2 1 2 0 0 1 2 2 3 4 5 0 01 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 1 0 01 2 2 1 0 0 -1 -1output:Case 1 is a tree.Case 2 is not a tree.Case 3 is not a tree.Case 4 is not a tree.Case 5 is not a tree.Case 6 is not a tree.*/#include<math.h>#include<stdio.h>  #include<string.h>#define inf 0x3ffffff#define maxn 10+100000#define max(a,b) ((a)>(b)?(a):(b))int N,M,K;int parent[maxn];int vis[maxn];int find(int *parent,int k){    if(parent[k]==-1)        return k;    parent[k]=find(parent,parent[k]);    return parent[k];}int main(){    int i,j,again;    int e,r,ans,ee,rr,cnt;int nn;    memset(parent,-1,sizeof(parent));    memset(vis,0,sizeof(vis));    nn=ans=again=cnt=0;    while(1)    {        if(again)        {            for(j=0,i=1;i<=nn;i++)            {                if(vis[i] && parent[i]==-1)                    j++;                if(j>1)                {ans=1;break;}            }if(nn==0)ans=0;            if(ans)                printf("Case %d is not a tree.\n",++cnt);                        else                printf("Case %d is a tree.\n",++cnt);            ans=again=0;            memset(parent,-1,sizeof(parent));            memset(vis,0,sizeof(vis));        }        scanf("%d%d",&e,&r);nn=max(nn,max(e,r));        if(e<0 && r<0)            break;        if(e==0  && r==0 )        {again=1;continue;}vis[r]=vis[e]=1;        if(ans==1)            continue;        ee=find(parent,e);        rr=find(parent,r);        if(r!=rr || rr==ee)            ans=1;        parent[rr]=ee;                }   return 0;  }

HDU 1325 poj 1308 is it a tree? (Query set)