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我的路径压缩函数int father[maxn],r[maxn]; //r数组记录当前节点和其父亲的关系 int findfather(int x,int &sum){ if (father[x]!=x){ father[x]=findfather(father[x],sum); sum=(sum+r[x])%3; } r[x]=sum; return father[x]; } r初始化皆为0; 传参:fx=findfather(x,tem=0); Followed by: Post your reply here: |
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