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

传送门

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
,b
,c
,ia
,ib
,ic
;

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
,stkb
,stkc
,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;
}