poj3257

dp，先将材料按以终点为关键字升序排

设f[i,j]为过山车到建到位置i在用了j元钱所得到的最大价值，然后

 var x,y,v,w:array[..] of longint;
     f:array[..,..] of longint;
     l,n,k,m,j,i,ans:longint;

 function max(a,b:longint):longint;
   begin
     if a>b then exit(a) else exit(b);
   end;

 procedure swap(var a,b:longint);
   var c:longint;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;

 procedure sort(l,r: longint);
   var i,j,p: longint;
   begin
     i:=l;
     j:=r;
     p:=y[(l+r) div ];
     repeat
       while y[i]<p do inc(i);
       while p<y[j] do dec(j);
       if not(i>j) then
       begin
         swap(x[i],x[j]);
         swap(y[i],y[j]);
         swap(v[i],v[j]);
         swap(w[i],w[j]);
         inc(i);
         j:=j-;
       end;
     until i>j;
     if l<j then sort(l,j);
     if i<r then sort(i,r);
   end;

 begin
   readln(l,n,m);
   for i:= to n do
   begin
     readln(x[i],j,v[i],w[i]);
     y[i]:=x[i]+j;
     if y[i]>l then y[i]:=l;
   end;
   sort(,n);
   j:=;
   fillchar(f,sizeof(f),);
   f[,]:=;
   for i:= to l do
   begin
     while y[j]<=i do
     begin
       if y[j]=i then
       begin
         for k:= to m do
           if (f[x[j],k]>) and (k+w[j]<=m) then
             f[i,k+w[j]]:=max(f[i,k+w[j]],f[x[j],k]+v[j]);
       end;
       inc(j);
     end;
   end;
   ans:=;
   for j:= to m do
     ans:=max(ans,f[l,j]);
   writeln(ans-);
 end.

容易分析复杂度为O(nm)

首尾必须相连是这题关键