HDU1272&POJ

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.
hdu1272和poj1308的题目是一样的，不过输出结果不同。正赶上今天poj崩了，只在hdu上a了这道题。
这道题的问形成的图是不是一棵树。树的要求有：
1.入度为1（根除外），出度不限。因为我是用并查集的思想做的，所以一发现par所指的节点想要改变就说明入度大于1了。
2.没有环，不形成森林。在进行 并 的过程中，记录边数和节点数，看是否相差1.
3.一个点或者没有点也是树。（没有点也是树？orz）
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下面的代码是过hdu的，用注释的替换原来的就能a了poj的

#include<iostream>#include<string>#include<algorithm>#include<stdlib.h>#include<stdio.h>#include<queue>#include<vector>using namespace std;#define N 100000+20#define INF 0x3f3f3f3f#define mem(arr,a) memset(arr,a,sizeof(arr))/*********************************/int par[N];int used[N];int n, m;int flag;int e;void unite(int a, int b){    e++;    if (par[b] == a){        flag = 1;        return;    }    par[b] = a;}int main(){    int cnt = 1;    int num = 0;    const char*no = "No";//"is not a tree.";    const char*yes = "Yes";// "is a tree.";    while (cin >> n >> m){        mem(used, 0);        mem(par, -1);        flag = 0;        e = 0;        num = 0;        if (n == -1 && m == -1)return 0;//      printf("Case %d ", cnt++);        if (n == 0 && m == 0){            cout << yes << endl;            continue;        }        num = 2;        par[n] = n;        unite(n, m);        while (cin >> n >> m){            if (n == 0 && m == 0){                if (!flag&&e == num - 1)cout << yes << endl;                else cout << no << endl;                break;            }            if (par[n] == -1){                par[n] = n;                num++;            }            if (par[m] == -1||par[m]==m){                if (par[m]==-1)                num++;                unite(n, m);            }            else{                flag = 1;                continue;            }        }    }}