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很明显是一道RMQ问题,倍增法,维护一下区域的最大/小值就行了。
1 var n,i,j,q,f,t,times:longint;
2 hmin,hmax:array[0..100,0..50000] of longint;
3 function min(a,b:longint):longint;
4 begin
5 if a>b then exit(b)
6 else exit(a);
7 end;
8 function max(a,b:longint):longint;
9 begin
10 if a>b then exit(a)
11 else exit(b);
12 end;
13 begin
14 readln(n,q);
15 for i:=1 to n do
16 begin
17 readln(hmax[0,i]);
18 hmin[0,i]:=hmax[0,i];
19 end;
20 for i:=1 to trunc(ln(n)/ln(2))+1 do
21 for j:=1 to n-(1 shl i)+1 do
22 begin
23 hmin[i,j]:=min(hmin[i-1,j],hmin[i-1,j+(1 shl (i-1))]);
24 hmax[i,j]:=max(hmax[i-1,j],hmax[i-1,j+(1 shl (i-1))]);
25 end;
26 for i:=1 to q do
27 begin
28 readln(f,t);
29 times:=trunc(ln(t-f+1)/ln(2));
30 writeln(max(hmax[times,f],hmax[times,t-(1 shl times)+1])-
31 min(hmin[times,f],hmin[times,t-(1 shl times)+1]));
32 end;
33 end.
P1699: [Usaco2007 Jan]Balanced Lineup排队
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原文地址:http://www.cnblogs.com/Kalenda/p/4818724.html