NOIP2006T2 金明的预算方案

这道题可以用树形DP，也可以直接01背包，因为每个物品的附件不能再作为主件，所以选取每个物品就只有这几种可能的情况：不选这个物品、选取这个物品、选取这个物品及它的一个附件、选取这个物品及它的两个附件。很容易地想出来这是个背包问题。

用f[i,j]表示前i件物品用了j元钱获得的最大价值（即重要度与价值的成绩最大），那么状态转移方程是

f[i,j]=max{

f[i-1,j],//不选这件物品

f[i-1,j-cost[i,0]]+cost[i,0]*import[i,0],//只选择主件

f[i-1,j-cost[i,0]-cost[i,1]]+cost[i,0]*import[i,0]+cost[i,1]*import[i,1],//选择主件和附件1

f[i-1,j-cost[i,0]-cost[i,2]]+cost[i,0]*import[i,0]+cost[i,2]*import[i,2],//选择主件和附件2

f[i-1,j-cost[i,0]-cost[i,1]-cost[i,2]]+cost[i,0]*import[i,0]+cost[i,1]*import[i,1]+cost[i,2]*import[i,2],//选择主件和附件1、附件2

}

注意其中第二维的边界。

 

[pascal 代码]

VAR
        a,b:array[1..600,0..3]of longint;
        f:array[0..61,0..32001]of longint;
        n,m,tot:longint;
Procedure init;
 var
        i,j,v,p,q:longint;
 begin
        readln(n,m);
        n:=n div 10;
        tot:=0;
        for i:=1 to m do
                 begin
                        readln(v,p,q);
                        v:=v div 10;
                        if q=0 then
                                begin
                                        inc(tot);
                                        a[tot,0]:=v;
                                        b[tot,0]:=p;
                                        a[tot,3]:=i;
                                end
                        else
                                begin
                                        for j:=1 to tot do if a[j,3]=q then break;
                                        if a[j,1]=0 then
                                                begin
                                                        a[j,1]:=v;b[j,1]:=p;
                                                end
                                        else
                                                begin
                                                        a[j,2]:=v;b[j,2]:=p;
                                                end;
                                end;
                 end;
 end;
Procedure dp;
 var
        i,j,k:longint;
 begin
        fillchar(f,sizeof(f),0);
        for i:=1 to tot do
                for j:=0 to n do
                        begin
                                f[i,j]:=f[i-1,j];
                                if (j-a[i,0]>=0) and (f[i-1,j-a[i,0]]+a[i,0]*b[i,0]>f[i,j]) then f[i,j]:=f[i-1,j-a[i,0]]+a[i,0]*b[i,0];
                                if (j-a[i,0]-a[i,1]>=0)and (f[i-1,j-a[i,0]-a[i,1]]+a[i,0]*b[i,0]+a[i,1]*b[i,1]>f[i,j]) then f[i,j]:=f[i-1,j-a[i,0]-a[i,1]]+a[i,0]*b[i,0]+a[i,1]*b[i,1];
                                if (j-a[i,0]-a[i,2]>=0)and (f[i-1,j-a[i,0]-a[i,2]]+a[i,0]*b[i,0]+a[i,2]*b[i,2]>f[i,j]) then f[i,j]:=f[i-1,j-a[i,0]-a[i,2]]+a[i,0]*b[i,0]+a[i,2]*b[i,2];
                                if (j-a[i,0]-a[i,1]-a[i,2]>=0) then
                                        if f[i-1,j-a[i,0]-a[i,1]-a[i,2]]+a[i,0]*b[i,0]+a[i,1]*b[i,1]+a[i,2]*b[i,2]>f[i,j] then f[i,j]:=f[i-1,j-a[i,0]-a[i,1]-a[i,2]]+a[i,0]*b[i,0]+a[i,1]*b[i,1]+a[i,2]*b[i,2];
                        end;
 end;
Begin
        init;
        dp;
        writeln(f[tot,n]*10);
End.