poj1308 （并查集）Is It A Tree?

题目链接：poj1308

题目的要求是让判断给出的点能否连接成一颗树，一棵树的话就只有一个树根，同时一个子节点不能同时拥有两个父亲，解法是用并查集，用father数组来记录每个节点的父节点，v数组来记录是否输入了节点。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>using namespace std;int father[100],v[100];void set(){    for(int i = 0; i < 100; i ++)    father[i] = i, v[i] = 0;}int find_father(int x){    if(father[x] != x)    {        father[x] = find_father(father[x]);    }    return father[x];}int main(){    int i,x,y,p = 1;    while(scanf("%d%d",&x,&y), x !=-1 && y != -1)    {        if(!x && !y)//空树也是树，真心坑人。。        {            printf("Case %d is a tree.\n",p++);            continue;        }        set();        father[y] = x;        v[x] = v[y] = 1;        int flag = 1;        if(x == y) flag = 0;//两个数相等表示同一点指向自己，不是树        while(scanf("%d%d",&x,&y),x&&y)        {            if(!flag) continue;            v[x] = v[y] = 1;            if(father[y] != y) flag = 0;//表示该节点已经有父节点了            int fun = find_father(x);            if(fun == y) flag = 0;//如果x的父节点是y，不能构成树，列如1 2 2 1 0 0            else father[y] = fun;        }        if(!flag)        {            printf("Case %d is not a tree.\n",p++);            continue;        }        int num = 0;        for(i = 0; i < 100; i ++)        {            if(v[i] && father[i] == i)            {                num ++;//表示根的数量            }        }        if(num > 1)            printf("Case %d is not a tree.\n",p++);        else            printf("Case %d is a tree.\n",p++);    }    return 0;}