poj1182 food chain (species and collection)

http://poj.org/problem?id=1182

R[x] = 0 means x and father are homogeneous
R[X] = 1 x Eat father
R[X] = 2 means x is eaten by the father
Because there are only three species of animals, and three species of animals constitute a ring
So the relationship between animals can be deduced.

The relationship between C and A is (R[b] + r[c])%3; The relationship between B and father A can be 0,1,2 for the relationship between 0,1,2 C and Father B. Each kind of value, when you deduce on paper, you can get this relationship.

This relationship will be used to modify the r[c] when the path is compressed.

How to tell if D a B is a lie? When FA = = fb. It shows that both animals A and B are in a set. Then we can deduce the relationship between them.

We know two black lines, we ask for the Red Line 2, that is, the relationship between A and B. If you know the Red Line 1, then deduce the relationship between A and B, then the same as the above figure

Red Line 1 = (3-r[b])%3; This enumerates the values of r[b], and then pushes it to know.

So if (R[a] + 3-r[b])%3==d-1 that's the truth.

If FA!=FB. Then we're going to merge two sets.

。 The same reason as above, the first red Line 1, and then seek the Red Line 2, and then find the Red Line 3. Then merge two collections

1#include <stdio.h>2 intfather[55555],r[55555];3 4 voidInitintN)5 {6      for(intI=1; i<=n; ++i)7     {8Father[i] =i;9R[i] =0;Ten     } One } A  - intFindintx) - { the     if(X==father[x])returnFather[x]; -     intt =find (Father[x]); -R[X] = (R[father[x]] + r[x])%3; -FATHER[X] =T; +     returnFather[x]; -     /* + derivation of the relationship between path compression, r[father[x]] [father[x] and root node, r[x] is the relationship between X and Father[x] A     */ at } - intMain () - { -     intN,k,i,d,x,y,ans,fx,fy; -scanf"%d%d",&n,&k); -     { inAns =0; - init (n); to          for(i=0; i<k; ++i) +         { -scanf"%d%d%d",&d,&x,&y); the             if(X>n | | y>n | | (d==2&& x==y)) ans++; *             Else $             {Panax NotoginsengFX =find (x); -FY =find (y); the                 if(fx==fy) +                 { A                     if((r[x]+3-r[y])%3!=d-1) ans++; the                 } +                 Else -                 { $FATHER[FX] =fy; $                     //R[fy] = (r[x]-r[y]+d-1+3)%3; -R[FX] = (d1+R[Y]-R[X])%3; -                 } the             } -         }Wuyiprintf"%d\n", ans); the     } -     return 0; Wu}

poj1182 food chain (species and collection)