[NOIP2014] Joint weighted value

Main topic:

There is a tree that asks for the sum of the product of the distance of 2 points and the maximum value.

Ideas:

Enumerating each point, the points adjacent to each other are points with a distance of 2. The maximum value of this part is the product of the two points with the maximum point weight, and the square of each point minus the weights and squares of the points, so that each edge is run two times, so the complexity is O (n).

Using adjacency tables to store double arrays (no direction), of course, like the Cyk of the big God directly running without considering this.

Code:

adjacency table:

1#include <cstdio>2 Const intmo=10007, m=200008;3 intcnt,x,y,n,i,ans,tot,w[m],v[m<<1],last[m<<1],head[m<<1];4 5 voidAddintXintY) {v[++cnt]=y,last[cnt]=head[x],head[x]=CNT;}6 7 intMain ()8 {9scanf"%d",&n);Ten      for(i=1; i<n;i++) scanf ("%d%d",&x,&y), add (x, y), add (y,x); One      for(i=1; i<=n;i++) scanf ("%d",&w[i]); A      for(i=1; i<=n;i++) -     { -         intsum=0, max1=0, max2=0, J,o; the          for(j=head[i];j;j=Last[j]) -         { -o=W[v[j]]; -sum= (sum+o)%mo; +             if(O&GT;MAX1) max2=max1,max1=o; -             Else if(O&GT;MAX2) max2=o; +tot= (tot-o*o)%mo; A         } attot= (tot+sum*sum)%mo; -Sum=max1*Max2; -         if(Sum>ans) ans=sum; -     } -printf"%d%d\n", Ans,tot); -     return 0; in}

%%%cyk Great God:

1#include <iostream>2 using namespacestd;3 inti,n,maxx,ans,x[200005],y[200005],w[200005];4 intam[200005],am2[200005],s[200005],qs[200005];5 intMain ()6 {7Cin>>N;8      for(i=1; i<=n-1; i++) cin>>x[i]>>Y[i];9      for(i=1; i<=n;i++) cin>>W[i];Ten      for(i=1; i<=n-1; i++) One       { AS[x[i]]= (S[x[i]]+w[y[i])%10007; -Qs[x[i]]= (Qs[x[i]]+w[y[i]]*w[y[i])%10007; -S[y[i]]= (S[y[i]]+w[x[i])%10007; theQs[y[i]]= (Qs[y[i]]+w[x[i]]*w[x[i])%10007; -           if(W[y[i]]>am[x[i]]) {am2[x[i]]=am[x[i]];am[x[i]]=W[y[i]];} -           Else if(W[y[i]]>am2[x[i]]) am2[x[i]]=W[y[i]]; -           if(W[x[i]]>am[y[i]]) {am2[y[i]]=am[y[i]];am[y[i]]=W[x[i]];} +           Else if(W[x[i]]>am2[y[i]]) am2[y[i]]=W[x[i]]; -       } +maxx=0; A      for(i=1; i<=n;i++) Maxx=max (maxx,am[i]*am2[i]); atans=0; -      for(i=1; i<=n;i++) ans= (Ans+s[i]*s[i]-qs[i])%10007; -cout<<maxx<<' '<<ans; -}

[NOIP2014] Joint weighted value