POJ 1182 食物链 *

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题目地址

用relaton[]数组保存结点与父节点的关系

同时要考虑两个集合并的时候，在状态压缩时要判断子结点与新的父节点的关系

思路来源：

http://blog.csdn.net/qq_33901573/article/details/52005905

http://blog.csdn.net/roney_win/article/details/9408571

#include<iostream>#include<cstdio>#include<queue>#include<map> #include<vector>#include<cctype>#include<cstring>#include<algorithm>using namespace std;const int maxn=50000+5;int p[maxn],r[maxn];//relationint getParent(int x){if(p[x]==x) return x;int xp=getParent(p[x]);r[x]=(r[x]+r[p[x]])%3;return p[x]=xp;}void merge(int d,int xp,int yp,int x,int y){p[yp]=xp;r[yp]=(3+(d-1)+r[x]-r[y])%3;}int main(){int N,K,D,X,Y;cin>>N>>K;int wrong=0;for(int i=1;i<=N;i++) p[i]=i;while(K--){cin>>D>>X>>Y;if(X>N||Y>N) wrong++;else if(D==2&&X==Y) wrong++;else {int xp=getParent(X);int yp=getParent(Y);if(xp!=yp) merge(D,xp,yp,X,Y);else{if(D==1&&r[X]!=r[Y]) wrong++;else if(D==2&&((r[Y]-r[X]+3)%3)!=1) wrong++;}}}cout<<wrong;return 0;}

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