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P1699: [Usaco2007 Jan]Balanced Lineup排队

时间:2015-09-18 13:27:32      阅读:155      评论:0      收藏:0      [点我收藏+]

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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

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