BZOJ3611: [Heoi2014] Big project

Exercises

After the consumption war, the difficulty of this problem is only in the statistical ans.

I originally asked for the longest (short) chain will only retain the suboptimal value, and then open three arrays to write a particularly troublesome ...

Studied today, Orz POPOQQQ.

InlinevoidDfsintx) {f[x]=v[x];g[x]=0; MI[X]=V[X]?0: INF; MX[X]=V[X]?0:-inf;            For4 (i,x) {dfs (y); Ans+ = (G[X]+F[X]*E[I].W) *f[y]+g[y]*F[x]; F[X]+=F[y]; G[X]+=g[y]+ (LL) e[i].w*F[y]; Ans1=min (ans1,mi[x]+mi[y]+E[I].W); Ans2=max (ans2,mx[x]+mx[y]+E[I].W); MI[X]=min (mi[x],mi[y]+E[I].W); MX[X]=max (mx[x],mx[y]+E[I].W); } Head[x]=0; }    

Instead, the LCA of the endpoint of the longest chain is enumerated directly, and the "prefix and" nature of the MI,MX array is fully exploited. Orzzzzzzzzzzzzz

Then the sum of the statistics is very simple, divided into two parts of the contribution can be.

Code:

1#include <cstdio>2#include <cstdlib>3#include <cmath>4#include <cstring>5#include <algorithm>6#include <iostream>7#include <vector>8#include <map>9#include <Set>Ten#include <queue> One#include <string> A #defineINF 1000000000 - #defineMAXN 1000000+5 - #defineMAXM 100000+5 the #defineEPS 1e-10 - #definell Long Long - #definePA pair<int,int> - #defineFor0 (i,n) for (int i=0;i<= (n); i++) + #defineFor1 (i,n) for (int i=1;i<= (n); i++) - #defineFor2 (i,x,y) for (int i= (x); i<= (y); i++) + #defineFor3 (i,x,y) for (int i= (x); i>= (y); i--) A #defineFor4 (i,x) for (int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go) at #defineMoD 1000000007 - using namespacestd; -InlineintRead () - { -     intx=0, f=1;CharCh=GetChar (); -      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} in      while(ch>='0'&&ch<='9') {x=Ten*x+ch-'0'; ch=GetChar ();} -     returnx*F; to } + intn,m,id[maxn],cnt,dep[maxn],fa[maxn][ +],F[MAXN],MI[MAXN],MX[MAXN]; - ll G[MAXN]; the BOOLV[MAXN]; * intAns1,ans2; $ ll ans;Panax NotoginsengInlineintLcaintXinty) - { the     if(dep[x]<Dep[y]) swap (x, y); +     intt=dep[x]-Dep[y]; AFor0 (I, -)if(t& (1<<i)) x=Fa[x][i]; the     if(x==y)returnx; +For3 (I, -,0)if(Fa[x][i]!=fa[y][i]) x=fa[x][i],y=Fa[y][i]; -     returnfa[x][0]; $ } $ structGraph - { -     intHead[maxn],tot; the     structedge{intGo,next,w;} e[2*MAXN]; -InlinevoidAddintXinty)Wuyi     { theE[++tot]= (Edge) {y,head[x],0};head[x]=tot; -E[++tot]= (Edge) {X,head[y],0};head[y]=tot; Wu     } -InlinevoidADDD (intXinty) About     { $E[++tot]= (Edge) {y,head[x],dep[y]-dep[x]};head[x]=tot; -     } -InlinevoidDfsintXintf) -     { Aid[x]=++CNT; +For1 (I, -) the         if(dep[x]>=1<<i) fa[x][i]=fa[fa[x][i-1]][i-1]; -         Else  Break; $For4 (i,x)if(y!=f) the         { thedep[y]=dep[x]+1; fa[y][0]=x; the DFS (y,x); the         } -     } inInlinevoidDfsintx) the     { thef[x]=v[x];g[x]=0; AboutMI[X]=V[X]?0: INF; theMX[X]=V[X]?0:-inf; the For4 (i,x) the         { + dfs (y); -ans+= (G[X]+F[X]*E[I].W) *f[y]+g[y]*F[x]; thef[x]+=F[y];Bayig[x]+=g[y]+ (LL) e[i].w*F[y]; theAns1=min (ans1,mi[x]+mi[y]+E[I].W); theAns2=max (ans2,mx[x]+mx[y]+E[I].W); -Mi[x]=min (mi[x],mi[y]+E[I].W); -Mx[x]=max (mx[x],mx[y]+E[I].W); the         } thehead[x]=0; the     }     the }g1,g2;  - intA[maxn],sta[maxn],top; theInlineBOOLcmpintXintY) {returnid[x]<id[y];}  the intMain () the {94Freopen ("Input.txt","R", stdin); theFreopen ("output.txt","W", stdout); then=read (); theFor1 (i,n-1) G1.add (read (), read ());98G1.dfs (1,0); About     intt=read (); -      while(t--)101     {102M=read (); ans=0; ans1=inf;ans2=-inf;103For1 (i,m) a[i]=read ();104Sort (A +1, a+m+1, CMP); theFor1 (i,m) v[a[i]]=1;106sta[top=1]=1; g2.tot=0;107 For1 (i,m)108         {109             intx=a[i],f=LCA (X,sta[top]); the              while(dep[f]<Dep[sta[top]])111             { the               if(dep[f]>=dep[sta[top-1]])113               { theG2.ADDD (f,sta[top--]); the                    if(sta[top]!=f) sta[++top]=F; the                     Break;117               }118G2.ADDD (sta[top-1],sta[top]); top--;119             } -             if(sta[top]!=x) sta[++top]=x;121         }122          while(--top) G2.ADDD (sta[top],sta[top+1]);123G2.dfs (1);124printf"%lld%d%d\n", ans,ans1,ans2); theFor1 (i,m) v[a[i]]=0;126     }        127     return 0; -}

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BZOJ3611: [Heoi2014] Big project