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[0..10010] of longint;
    f:array[0..1010,0..1010] 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 2];
    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-1;
      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:=1 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(1,n);
  j:=1;
  fillchar(f,sizeof(f),0);
  f[0,0]:=1;
  for i:=1 to l do
  begin
    while y[j]<=i do
    begin
      if y[j]=i then
      begin
        for k:=0 to m do
          if (f[x[j],k]>0) 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:=0;
  for j:=1 to m do
    ans:=max(ans,f[l,j]);
  writeln(ans-1);
end.

容易分析复杂度为O(nm)

首尾必须相连是这题关键