HDU ACM 1325 / POJ 1308 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): 6961    Accepted Submission(s): 1619

【题目链接】http://acm.hdu.edu.cn/showproblem.php?pid=1325

这题如果用并查集做的话，最后判断的时候还是要看一个结点的入度是否为大于1（纯并查集的情况下），搜了几份代码，有些是去判断是否有环的，这些都不如直接用树的充分条件去判断：一个节点入度为0，其余节点入读为1，V个点只有V-1条边，题目坑你的地方大概有两点：没节点也是树，结束时两个数是负数而不是两个-1

 #include <cstdio>
 #include <cstring>
 #define SIZE 2013
 using namespace std;
 int father[SIZE];
 bool exit[SIZE];
 int find(int f)
 {
     return father[f] = f == father[f] ? f : find(father[f]);
 }

 int main()
 {
     #ifndef ONLINE_JUDGE
     freopen("input.txt", "r", stdin);
     #endif
     int n, m, T = , max = , nv = , ne = ;
     bool flag = true;
     memset(father, , sizeof(father));
     memset(exit, false, sizeof(exit));
     while(scanf("%d%d", &n, &m) != EOF)
     {
         if(n <  && m < ) break;
         else if(n+m == )
         {
             if(ne ==  && nv == )
             {
                     printf("Case %d is a tree.\n", ++T);
                     continue;
             }
             if(ne == nv-)
             {
                 int cnt = , zero = ;
                 for(int i = ; i <= max; ++i)
                 if(exit[i])
                 {
                     if(!father[i]) zero++;
                     else if(father[i] == ) cnt++;
                 }
                 if(cnt == ne && zero == )
                     printf("Case %d is a tree.\n", ++T);
                 else
                     printf("Case %d is not a tree.\n", ++T);
             }
             else printf("Case %d is not a tree.\n", ++T);

             flag = true;
             memset(exit, false, sizeof(exit));
             memset(father, , sizeof(father));
             ne = nv = ;
             max = ;
         }
         else
         {
             max = max > n ? (max > m ? max : m) : (n > m ? n : m);
             if(!exit[n])
             {
                 exit[n] = true;
                 nv++;
             }
             if(!exit[m])
             {
                 exit[m] = true;
                 nv++;
             }
             ne++;
             father[m]++;
         }
     }
     return ;
 }