Meaning of the questions:
There are n matryoshka doll inner product of each outer product out in wasted space means a plurality of (or one) doll in combination with the maximum volume of air inside the doll
Q. How many combinations there are sets of baby (can not be caught apparently caught someone else sets her answer will not be the smallest) makes a minimum of wasted space
answer:
Segment tree optimization dp
- Provided dp [i] .minn denotes the i th doll as the innermost doll minimal wasted space doll all combinations it is clear that the state transferred from $ inj> outi $ in
- N in accordance with a first set of her inner product ordering again easy to find nested object that is j
- First, obviously binary search backwards dp j satisfy the minimum $ inj> out $ jn the doll so meet the minimum transition condition then direct the search in the i-th position of the insertion $ dp [j] .minn- (out [i] -in [i]), dp [j] .num $ to
- The tree-like structure with an array of convenient and a lot can be optimized push_up to merge (after teammate remind tql !!!)
#include<bits/stdc++.h> using namespace std; const int N=2e5+10; int a[N],num,minn; #define ll long long ll ans; const ll mod=1e9+7; struct node{ll minn,num;}t[N<<2]; node merge(node a,node b) { if(a.minn==b.minn)return (node){a.minn,(a.num+b.num)%mod }; if(a.minn<b.minn) return (node){a.minn,a.num}; if(a.minn>b.minn) return (node){b.minn,b.num}; } void upnode(int x,int v,int num,int l,int r,int pos) { if(l==r){t[pos]=(node){v,num};return;} int m=(l+r)>>1; if(x<=m)upnode(x,v,num,l,m,pos<<1); else upnode(x,v,num,m+1,r,pos<<1|1); t[pos]=merge(t[pos<<1],t[pos<<1|1]); } node qmin(int L,int R,int l,int r,int pos) { if(L<=l&&r<=R)return t[pos]; int m=l+r>>1;node ans=(node){1000000000,0}; if(L<=m)ans=merge(ans,qmin(L,R,l,m,pos<<1)); if(R>m)ans=merge(ans,qmin(L,R,m+1,r,pos<<1|1)); return ans; } struct NODE{int in,out;}s[N]; int n; int main() { cin>>n; for(int i=1;i<=n;i++)scanf("%d%d",&s[i].out,&s[i].in); sort(s+1,s+1+n,[](NODE a,NODE b){return a.in<b.in;}); for(int i=n;i>=1;i--) { int L=1,R=n,ans=-1; while(L<=R) { int mid=(L+R)>>1; if(s[mid].in>=s[i].out)ans=mid,R=mid-1; else L=mid+1; } if(ans==-1)upnode(i,s[i].in,1,1,n,1); else { node u=qmin(ans,n,1,n,1); upnode(i,u.minn-(s[i].out-s[i].in),u.num,1,n,1); } } cout<<t[1].num; return 0; }
Shortest count
- We want to maintain what is $ in [pk] + (in [pk-1] -out [pk-1]) + (in [pk-2] - out [pk-1]) + .... + ( in [pk1] -out [pk1]) $ and the minimum number satisfying $ out [pk]> in [n] $
- We can even from FIG. $ 0-n-1 add (i, i + 1, in [i + 1] -in [i]) $ run, then maintained under this path is in [pk]
- There are (in [pk-1] - out [pk-1]) can be optimized shortest set the Add Minimum temp $ object may be transferred (i, temp, in [temp] -out [i]) $ can be found just run out of that one above formula
#include<bits/stdc++.h> using namespace std; const int N=2e5+10; const int mod=1e9+7; struct node{int in,out;}s[N]; int n,ans,dis[N],cnt[N],vis[N],pos,head[N],minn=1e9; struct Edge{int to,nex,v;}edge[N<<1]; void add(int a,int b,int c){edge[++pos]=(Edge){b,head[a],c};head[a]=pos;}; int main() { cin>>n; for(int i=1;i<=n;i++)scanf("%d%d",&s[i].out,&s[i].in); sort(s+1,s+1+n,[](node a,node b){return a.in<b.in;}); for(int i=0;i<n;i++)add(i,i+1,s[i+1].in-s[i].in); for(int i=1;i<=n;i++) { int temp=-1,L=1,R=n; while(L<=R) { int mid=L+R>>1; if(s[mid].in>=s[i].out)temp=mid,R=mid-1;else L=mid+1; } if(temp==-1)continue; else add(i,temp,s[temp].in-s[i].out); } memset(dis,0x3f,sizeof dis); dis[0]=0;cnt[0]=1; for(int i=0;i<=n;i++) { for(int j=head[i];j;j=edge[j].nex) { int v=edge[j].to; if(dis[i]+edge[j].v<dis[v])dis[v]=dis[i]+edge[j].v,cnt[v]=cnt[i]; else if(dis[i]+edge[j].v==dis[v])cnt[v]+=cnt[i],cnt[v]%=mod; } } for(int i=1;i<=n;i++{if(s[i].out>s[n].in&&dis[i]<minn)minn=dis[i];} for(int i=1;i<=n;i++) if(s[i].out>s[n].in&&dis[i]==minn)ans=(ans+cnt[i])%mod; cout<<ans; return 0; }