HDOJ1325

HDOJ1325
乍一看并查集啊，native，结果WA, ORZ
还是太菜了啊，做了好久好久orz

/*三个条件：1.不能成环2.不能成树林3.入度小于等于1特判：空树也是树*/#include<iostream>#include<stdio.h>using namespace std;#define MAXN 10005int disjoint[MAXN];   //并查集int indegree[MAXN];   //记录出度int show[MAXN];        //标记出现的点int find(int x)       //找根节点{    int t = x;    while(disjoint[t] != t)        t = disjoint[t];    return t;}int merge(int x, int y)  //合并查集(如果在一个集合形成环，则返回0){    int fx = find(x);    int fy = find(y);    if (fx == fy) return 0;    else while(disjoint[fy]!= fx)        disjoint[fy] = fx;    return fx;//如果没有成环，则合并，否则返回0                                             }int main(){    int x, y,count=1;    while (cin >> x >> y&&x>=0&&y>=0)    {        for (int i = 1; i <= MAXN; i++)      //初始化并查集和入度和标记        {            indegree[i] = 0;            disjoint[i] = i;            show[i] = 0;        }        if (x == 0 && y == 0)                 //特判空树        {            printf("Case %d is a tree.\n", count++);            continue;        }        show[x]++; show[y]++;        indegree[y]++;            merge(x, y);                         //要注意x，y顺序，因为该树是有方向的        int flag1 = 0, flag2 = 0,flag3=0;    //分别表示是否成环，是否成树林，是否入度小于等于1        int a, b;        while (cin >> a >> b&&a&&b)        {            if (merge(a, b) == 0)flag1 = 1;//成环            show[a]++; show[b]++;            indegree[b]++;        }        int root=0;        if(flag1==1)printf("Case %d is not a tree.\n", count++);        else        {            for (int j = 1; j < MAXN; j++)            {                if (show[j] && disjoint[j] == j)  //计算有几个根节点                {                    root++;                    if (root > 1) {                        flag2 = 1;                        break;                    }                }                if (indegree[j] > 1)flag3 = 1;            }            if(flag2||flag3)printf("Case %d is not a tree.\n", count++);            else printf("Case %d is a tree.\n", count++);        }    }    return 0;}