[线段树][单调栈]HackerRank 101 Hack 50 .Boxes for Toys

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首先可以发现，将三元组排序，一个区间里的maxa∗maxb∗maxc即是这个区间的权值。

考虑枚举右端点r，用线段树维护每个i,i∈[1,r]，i到r中maxa∗maxb∗maxc的值，这样用单调栈维护a,b,c，退栈入栈分别用区间除，区间乘更新

#include <cstdio>#include <iostream>#include <algorithm>#include <ctime>using namespace std;const int N=300010,P=1e9+7;typedef long long ll;int n;int a[N],b[N],c[N],ia[N],ib[N],ic[N];int prod[N<<2],tag[N<<2];inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void rea(int &x){  char c=nc(); x=0;  for(;c>'9'||c<'0';c=nc());for(;c>='0'&&c<='9';x=x*10+c-'0',c=nc());}inline void add(int &x,int y){  (x+=y)%=P;}inline ll Pow(ll x,int y){  ll ret=1; x%=P;  for(;y;y>>=1,x=x*x%P) if(y&1) ret=ret*x%P;  return ret;}inline void Push(int g){  if(tag[g]!=1){    prod[g<<1]=1LL*prod[g<<1]*tag[g]%P;    tag[g<<1]=1LL*tag[g<<1]*tag[g]%P;    prod[g<<1|1]=1LL*prod[g<<1|1]*tag[g]%P;    tag[g<<1|1]=1LL*tag[g<<1|1]*tag[g]%P;    tag[g]=1;  }}inline void Up(int g){  prod[g]=(prod[g<<1]+prod[g<<1|1])%P;}void Build(int g,int l,int r){  tag[g]=1;  if(l==r) return prod[g]=1,void();  int mid=l+r>>1;  Build(g<<1,l,mid); Build(g<<1|1,mid+1,r);  Up(g);}void Modify(int g,int l,int r,int L,int R,int x){  if(l==L&&r==R){    tag[g]=1LL*tag[g]*x%P; prod[g]=1LL*prod[g]*x%P;    return ;  }  Push(g); int mid=L+R>>1;  if(r<=mid) Modify(g<<1,l,r,L,mid,x);  else if(l>mid) Modify(g<<1|1,l,r,mid+1,R,x);  else Modify(g<<1,l,mid,L,mid,x),Modify(g<<1|1,mid+1,r,mid+1,R,x);  Up(g);}int Query(int g,int l,int r,int L,int R){  if(l==L&&r==R) return prod[g];  Push(g); int mid=L+R>>1;  if(r<=mid) return Query(g<<1,l,r,L,mid);  if(l>mid) return Query(g<<1|1,l,r,mid+1,R);  return (Query(g<<1,l,mid,L,mid)+Query(g<<1|1,mid+1,r,mid+1,R))%P;}int stka[N],stkb[N],stkc[N],tpa,tpb,tpc;int main(){  freopen("box.in","r",stdin);  freopen("box.out","w",stdout);  rea(n);  for(int i=1;i<=n;i++){    rea(a[i]); rea(b[i]); rea(c[i]);    if(b[i]>a[i]) swap(a[i],b[i]);    if(c[i]>a[i]) swap(a[i],c[i]);    if(c[i]>b[i]) swap(b[i],c[i]);    ia[i]=Pow(a[i],P-2); ib[i]=Pow(b[i],P-2); ic[i]=Pow(c[i],P-2);  }  Build(1,1,n);  int ans=0;  for(int i=1;i<=n;i++){    while(tpa && a[i]>a[stka[tpa]])      Modify(1,stka[tpa-1]+1,stka[tpa],1,n,ia[stka[tpa]]),tpa--;    Modify(1,stka[tpa]+1,i,1,n,a[i]); stka[++tpa]=i;    while(tpb && b[i]>b[stkb[tpb]])      Modify(1,stkb[tpb-1]+1,stkb[tpb],1,n,ib[stkb[tpb]]),tpb--;    Modify(1,stkb[tpb]+1,i,1,n,b[i]); stkb[++tpb]=i;    while(tpc && c[i]>c[stkc[tpc]])      Modify(1,stkc[tpc-1]+1,stkc[tpc],1,n,ic[stkc[tpc]]),tpc--;    Modify(1,stkc[tpc]+1,i,1,n,c[i]); stkc[++tpc]=i;    add(ans,Query(1,1,i,1,n));  }  ans=1LL*ans*Pow(1LL*(n+1)*n/2,P-2)%P;  printf("%d\n",ans);  return 0;}