[CF915F]Imbalance Value of a Tree
2018-05-29 20:49
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[CF915F]Imbalance Value of a Tree
题目大意:
一棵\(n(n\le10^6)\)个结点的树,每个结点有一个权值\(w_i\)。定义\(I(i,j)\)为\(i\)到\(j\)之间简单路径上最大权值与最小权值之差,求\(\displaystyle\sum_{i=1}^n\sum_{j=1}^nI(i,j)\)。
思路:
分别计算路径最大权值之和与最小权值之和。以最大权值之和为例,在图中按权值从大到小枚举每一个点,则对于该连通块中每一个经过该点的路径,该点为路径上权值最大的点,可以计算该点对答案的贡献,并将计算完贡献的点从图中删去。
由于删点是一个难以实现的操作,因此可以将操作改为加点操作,用并查集维护连通性即可。最小权值和同理。时间复杂度\(\mathcal O(n\log n)\)。
源代码:
#include<cstdio> #include<cctype> #include<numeric> #include<algorithm> #include<forward_list> using int64=long long; inline int getint() { register char ch; while(!isdigit(ch=getchar())); register int x=ch^'0'; while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0'); return x; } constexpr int N=1e6+1; int seq ,pos ,w ; std::forward_list<int> e ; inline void add_edge(const int &u,const int &v) { e[u].emplace_front(v); e[v].emplace_front(u); } struct DisjointSet { int anc ,size ; void reset(const int &n) { std::fill(&size[1],&size[n+1],1); std::iota(&anc[1],&anc[n+1],1); } int find(const int &x) { return x==anc[x]?x:anc[x]=find(anc[x]); } void merge(const int &x,const int &y) { size[find(y)]+=size[find(x)]; anc[find(x)]=find(y); } }; DisjointSet s; int main() { const int n=getint(); for(register int i=1;i<=n;i++) w[i]=getint(); for(register int i=1;i<n;i++) { add_edge(getint(),getint()); } int64 max=0,min=0; s.reset(n); std::iota(&seq[1],&seq[n+1],1); std::sort(&seq[1],&seq[n+1],[](const int &a,const int &b){return w[a]<w[b];}); for(register int i=1;i<=n;i++) pos[seq[i]]=i; for(register int i=1;i<=n;i++) { const int &x=seq[i]; int64 last=1,tmp=0; for(register auto &y:e[x]) { if(pos[y]>pos[x]) continue; tmp+=last*s.size[s.find(y)]; last+=s.size[s.find(y)]; s.merge(x,y); } max+=(int64)w[x]*tmp; } s.reset(n); std::iota(&seq[1],&seq[n+1],1); std::sort(&seq[1],&seq[n+1],[](const int &a,const int &b){return w[a]>w[b];}); for(register int i=1;i<=n;i++) pos[seq[i]]=i; for(register int i=1;i<=n;i++) { const int &x=seq[i]; int64 last=1,tmp=0; for(register auto &y:e[x]) { if(pos[y]>pos[x]) continue; tmp+=last*s.size[s.find(y)]; last+=s.size[s.find(y)]; s.merge(x,y); } min+=(int64)w[x]*tmp; } printf("%lld\n",max-min); return 0; }
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