P1572: [Usaco2009 Open]工作安排Job
做这道题走了不少弯路，其实本身也是很简单的，类似单调队列的东西。刚开始以为双关键字排序就行了，结果连WA两遍，忽然意识到可以在截止之前做这件事！！于是就规规矩矩的打队列，然而忘记找队列里的最小P做，当成是单调的了，于是又顺顺利利地WA了好多次了，说多了都是泪啊。。
1 var n,i,j,l,r,min,mpos:longint;
2 ans:int64;
3 d,p,q:array[0..100000] of longint;
4 procedure swap(var a,b:longint);
5 var tem:longint;
6 begin
7 tem:=a;
8 a:=b;
9 b:=tem;
10 end;
11 procedure qs(t,w:longint);
12 var mid1,mid2,l,r:longint;
13 begin
14 l:=t; r:=w;
15 mid1:=d[(l+r) shr 1]; mid2:=p[(l+r) shr 1];
16 repeat
17 begin
18 while (d[l]<mid1) or ((d[l]=mid1) and (p[l]<mid2)) do inc(l);
19 while (d[r]>mid1) or ((d[r]=mid1) and (p[r]>mid2)) do dec(r);
20 if l<=r then
21 begin
22 swap(d[l],d[r]);
23 swap(p[l],p[r]);
24 inc(l); dec(r);
25 end;
26 end;
27 until l>r;
28 if t<r then qs(t,r);
29 if l<w then qs(l,w);
30 end;
31 begin
32 readln(n);
33 for i:=1 to n do
34 readln(d[i],p[i]);
35 qs(1,n);
36 {nowd:=-1;
37 for i:=1 to n do
38 begin
39 if nowd<>d[i] then
40 begin
41 nowd:=d[i];
42 inc(ans,p[i]);
43 end;
44 end;}
45 l:=1; r:=1; q[1]:=p[1];
46 for i:=2 to n do
47 begin
48 min:=maxlongint;
49 if r-l+1<d[i] then
50 begin
51 inc(r);
52 q[r]:=p[i];
53 end
54 else
55 begin
56 for j:=l to r do
57 if min>q[j] then
58 begin
59 min:=q[j];
60 mpos:=j;
61 end;
62 if p[i]>min then
63 q[mpos]:=p[i];
64 end;
65 end;
66 for i:=1 to r do inc(ans,q[i]);
67 writeln(ans);
68 end.
(转载请注明出处：http://www.cnblogs.com/Kalenda/)
