Topic Portal: http://poj.org/problem?id=1182
This is a relationship type and look up the problem, for each animal, there are only three cases: the same kind, eat and be eaten;
So you can use the 0,1,2 three numbers to represent three cases, in the use and check the set of the time to add a relational array, the initial value of all assigned to 0
Then it is the change of the relationship that is added to each step of the set,
If the ancestor node is the same, the relationship between the two has already appeared, is known, and determines whether the relationship is the same as the given d-1
The ancestor nodes are different, which means that the relationship between the two is unknown before the relationship is given.
Oh, yes, on the POJ. A single set of inputs, multiple groups will WA
Code
1#include <cstdio>2 using namespacestd;3 intfather[50005];4 intrelat[50005];5 voidGive (intN)6 {7 for(intI=1; i<=n;i++)8 {9father[i]=i;Tenrelat[i]=0;//initial relationship is 0 One } A } - intFindintx) - { the if(X==father[x])returnFather[x]; - intt=find (Father[x]); -Relat[x]= (Relat[x]+relat[father[x])%3and//To update the relationship during the search for the ancestor node -father[x]=T; + returnFather[x]; - } + intMain () A { at intN,k,sum,d,x,y,sx,sy; -scanf"%d%d",&n,&k); - give (n); -sum=0; - while(k--) - { inscanf" %d%d%d",&d,&x,&y); - if(x>n| | Y>N) to { +sum++; - Continue; the } * if(d==2&&x==y)//You can't eat yourself. $ {Panax Notoginsengsum++; - Continue; the } +sx=find (x); Asy=find (y); the if(sx==sy)//Description relationship can be determined because the previous occurrence of + { - if((relat[x]-relat[y]+3)%3!=d-1)//advocated to avoid negative numbers $sum++; $ } - else//relationship is uncertain, then can not judge, can only give the relationship - { thefather[sx]=Sy; -Relat[sx]= (relat[y]-relat[x]+3+d-1)%3; The order of Sx,sy,x,y is not reversed, this is the correspondingWuyi } the } -printf"%d\n", sum); Wu return 0; -}
POJ 1182 (relationship and search) food chain