[HAOI2015] Tree dyed (tree backpack)

There is a tree with a number of N, and the edge of the tree is right. Give you a positive integer k within 0~ N, you want to select K points in this tree, dye it black, and turn the other n-k Dianran into white. After all points are dyed, you will get the distance between the black dots 22 plus the white Point 22 distance and the benefit. Ask what the maximum benefit is.

Solution

Compare the classic tree knapsack problem.

If only the points are analyzed, the situation becomes very cumbersome and does not take into account each variable contribution, each edge produces a product of the number of black dots on both sides plus the product of the white dots on both sides.

In this way we can run the backpack directly, the standard tree-shaped backpack is n^3, but this problem each word backpack volume has an upper limit, the total complexity can be done n^2.

Code

#include <iostream>#include<cstdio>#include<cstring>#defineN 2009using namespacestd;Long LongDp[n][n];intSize[n],m,n,a,b,c,tot,head[n];structdsd{intn,to,l;} An[n<<1];inlinevoidAddintUintVintl) {an[++tot].n=Head[u]; An[tot].to=v; Head[u]=tot; AN[TOT].L=l;}voidDfsintUintFA) {Size[u]=1;  for(intI=head[u];i;i=AN[I].N)if(an[i].to!=FA) {      intv=an[i].to;    DFS (V,U); Size[u]+=Size[v];  for(intJ=min (M,size[u]); j>=0;--j)//       for(intk=0; K<=min (J,size[v]); + +k)if(dp[v][k]!=-0x3f3f3f3f)      {          Long LongNum= (Long Long) (k* (m-k) + (n-size[v]-(m-k)) * (SIZE[V]-K)) *AN[I].L; DP[U][J]=max (Dp[u][j],dp[u][j-k]+dp[v][k]+num); }  } }intMain () {scanf ("%d%d",&n,&m);  for(intI=1; i<n;++i) {scanf ("%d%d%d",&a,&b,&B);    Add (a,b,c); add (b,a,c); } memset (DP,-0x3f,sizeof(DP));  for(intI=1; i<=n;++i) dp[i][0]=dp[i][1]=0;//Dfs1,0); cout<<dp[1][m]; return 0;} 

This writing is too slow, and did not do strictly N^2,bzoj will tle, the following writing is stable.

Code

#include <iostream>#include<cstdio>#include<cstring>#defineN 2009using namespaceStd;typedefLong LongLl;ll Dp[n][n],size[n],m,n,a,b,c,tot,head[n],g[n];structdsd{ll N,to,l;} An[n<<1];inlinevoidAdd (ll u,ll v,ll l) {an[++tot].n=Head[u]; An[tot].to=v; Head[u]=tot; AN[TOT].L=l;} ll mi (ll X,ll y) {returnX<y?x:y;} ll Ma (ll x,ll y) {returnX<y?y:x;}voidDfs (ll u,ll FA) {Size[u]=1;  for(LL i=head[u];i;i=AN[I].N)if(an[i].to!=FA) {LL v=an[i].to;    DFS (V,U); ll x=mi (M,size[u]), y=mi (m,size[v]);  for(intj=0; j<=m;++j) g[j]=0;  for(LL j=x;j>=0;--j) for(intk=0; k<=y;++k)if(j+k<=m) {ll Gyx= (ll) k* (m-k) + (n-size[v]-(m-k)) * (SIZE[V]-K)) *AN[I].L; G[j+k]=ma (g[j+k],dp[u][j]+dp[v][k]+Gyx); }     for(intj=0; j<=m;++j) dp[u][j]=G[j]; Size[u]+=Size[v]; }}inlineintRd () {intx=0;CharC=GetChar ();  while(!isdigit (c)) c=GetChar ();  while(IsDigit (c)) {x= (x<<1) + (x<<3) + (c^ -); C=GetChar (); }    returnx;} intMain () {n=rd (); m=Rd ();  for(intI=1; i<n;++i) {a=rd (); B=rd (); c=Rd ();    Add (a,b,c); add (b,a,c); } memset (DP,-0x3f,sizeof(DP));  for(intI=1; i<=n;++i) dp[i][0]=dp[i][1]=0; DFS (1,0); printf ("%lld", dp[1][m]); return 0;} 

[HAOI2015] Tree dyed (tree backpack)