POJ 1182 食物链

题目链接：http://poj.org/problem?id=1182

详细题解：http://blog.csdn.net/c0de4fun/article/details/7318642 （没有比这篇食物链的结题报告更好的了）

附上我的代码：

#include<cstdio>
#define N 55000
int F[N], rel[N];

int fun(int x)
{
    if(x != F[x])
    {
       int fa_x = fun(F[x]);
       rel[x] = (rel[x] + rel[F[x]]) % 3;
       F[x] = fa_x;
    }

    return F[x];
}

void Union(int x, int y, int fa_x, int fa_y,int d)
{
    F[fa_y] = fa_x;
    rel[fa_y] = (3-rel[y]+(d-1)+rel[x])%3;
}
int slove(int d, int x, int y)
{
    int fa_x = fun(x);
    int fa_y = fun(y);
    if(d == 1)
    {
        if(fa_x == fa_y)  //在同一个集合内
        {
            if(rel[x] != rel[y]) return 1;
            else return 0;
        }
        else{
            Union(x,y,fa_x,fa_y,d);
            return 0;
        }
    }
    if(d == 2)
    {
        if(fa_x == fa_y)
        {
            if((rel[y]+3-rel[x])%3 != 1) return 1;
            else return 0;
        }
        else
        {
             Union(x,y,fa_x,fa_y,d);
             return 0;
        }
    }
}
int main ()
{
    int n, k, d, x, y;
    scanf("%d %d", &n, &k);
    for(int i = 1; i <= n; i++)
        F[i] = i, rel[i] = 0;

    int ans = 0;
    while(k--)
    {
        scanf("%d %d %d", &d, &x, &y);
        if(x > n||y > n){ans++; continue;}
        if(d == 2 && x == y){ans++; continue;}

        ans += slove(d,x,y);
    }
    printf("%d\n", ans);
    return 0;
}