这里特别说明一下,配位堆是支持单点修改的,只不过大根堆只能增值,小根堆只能减值,由于要维护堆的性质不变,复杂度$O(1)$。想要修改为任意值也能够,能够先把要修改的点及其子树取出,并断开其与子树的连边,修改后再将 此点及其全部子树 与根合并,相似pop操做,复杂度大概是$O(logn)$?,谨慎使用为好。node
#include<iostream> #include<cstdio> #include<cstring> #include<queue> #define N 10010 using namespace std; template<class T,int maxsize,bool cmp(const T &a,const T &b)> class Pairing_Heap { private: T val[maxsize]; int head[maxsize],next[maxsize],to[maxsize],fa[maxsize],root,size,cnt; int epool[maxsize],npool[maxsize],e,n; int stack[maxsize],t; inline int new_edge(){return e?epool[e--]:++cnt;} inline int new_node(){return n?npool[n--]:++size;} inline void add(int a,int b){int u=new_edge();to[u]=b;next[u]=head[a];head[a]=u;} public: Pairing_Heap(){e=n=size=cnt=root=t=0;} inline int merge(int a,int b){if(cmp(val[b],val[a]))swap(a,b);add(fa[b]=a,b);return a;} inline void push(T v){int u=new_node();val[u]=v;root=root?merge(root,u):u;} inline T top(){return val[root];} inline bool empty(){return !root;} inline void pop() { register int i;t=0; for(i=head[root];i;i=next[i]){epool[++e]=i;if(fa[to[i]]==root)fa[s[++t]=to[i]]=0;} fa[root]=head[root]=0,npool[++n]=root,root=i=0; while(i<t){++i;if(i==t){root=s[i];return;}int u=s[i],v=s[++i];s[++t]=merge(u,v);} } }
直接根根合并,并使小根为大根子树(大根堆)
ios