PKU1170 shopping trip

       一道极丑的背包和冲进ranklist前10的代码……我们把每件商品和每种优惠方案都看做一个有一定价值和容量的物品，直接背包问题解决之！

program shopping_offers;const maxn=5; maxt=110; inf=10000000;type rec=record sum:array[0..maxn] of longint; cost:longint; end;var f:array[0..maxn,0..maxn,0..maxn,0..maxn,0..maxn] of longint; data:array[0..maxt] of rec; num,need:array[0..maxn] of longint; n,m,tot:longint;procedure prework;begin tot:=0; filldword(f,sizeof(f) shr 2,inf); fillchar(need,sizeof(need),0); f[0,0,0,0,0]:=0;end;function find(x:longint):longint;var i:longint;begin for i:=1 to n do if num[i]=x then exit(i);end;procedure init;var i,j,t,k,x,s:longint;begin readln(n); for i:=1 to n do begin read(num[i]); read(need[i]); inc(tot); with data[tot] do begin fillchar(sum,sizeof(sum),0); readln(cost); inc(sum[i]); end; end; readln(m); for i:=1 to m do begin inc(tot); read(t); with data[tot] do begin fillchar(sum,sizeof(sum),0); for j:=1 to t do begin read(x); k:=find(x); read(s); inc(sum[k],s); end; readln(cost); end; end;end;procedure relax(i,a,b,c,d,e:longint);var tmp:longint;begin with data[i] do tmp:=f[a-sum[1],b-sum[2],c-sum[3],d-sum[4],e-sum[5]]+cost; if tmp < f[a,b,c,d,e] then f[a,b,c,d,e]:=tmp;end;procedure dp;var i,a,b,c,d,e:longint;begin for i:=1 to tot do with data[i] do for a:=sum[1] to maxn do for b:=sum[2] to maxn do for c:=sum[3] to maxn do for d:=sum[4] to maxn do for e:=sum[5] to maxn do relax(i,a,b,c,d,e);end;procedure outit;begin writeln(f[need[1],need[2],need[3],need[4],need[5]]);end;begin assign(input,'a.in'); reset(input); assign(output,'a.out'); rewrite(output); prework; init; dp; outit; close(input); close(output);end.