本来想给出一个链接，tyvjp1039忠诚2，大家自己找一下吧。这个问题很明显是区间极值问题，而且是动态的，这样就不能用RMQ了。线段树是一个比较好的选择。而且我发现一个问题，从前以为线段树之类的东西代码比较死，基本上都一样，其实我错了，越是高级的数据结构写法往往越活，变通性很强。只可惜此生无缘noi啦，这能自己虐一些简单的题目找找爽啦！悲剧呀！
1 type
2 ji=^rec;
3 rec=record
4 l,r,min:longint;
5 lson,rson:ji;
6 end;
7
8 var
9 a:ji;
10 b:array[0..100000] of longint;
11 i,j,x,y,m,n:longint;
12
13 function minn(x,y:longint):longint;
14 begin
15 if x<y then exit(x);
16 exit(y);
17 end;
18
19 procedure build(var a:ji; l,r:longint);
20 var
21 mid:longint;
22 begin
23 new(a);
24 a^.l:=l;
25 a^.r:=r;
26 mid:=(l+r)>>1;
27 if l=r then
28 begin
29 a^.min:=b[l];
30 exit;
31 end;
32 build(a^.lson,l,mid);
33 build(a^.rson,mid+1,r);
34 a^.min:=minn(a^.lson^.min,a^.rson^.min);
35 end;
36
37 function find(a:ji; l,r:longint):longint;
38 var
39 mid,k:longint;
40 begin
41 if (l<=a^.l)and(r>=a^.r) then exit(a^.min);
42 mid:=(a^.l+a^.r)>>1; k:=maxlongint;
43 if mid>=l then k:=minn(k,find(a^.lson,l,r));
44 if mid<r then k:=minn(k,find(a^.rson,l,r));
45 exit(k);
46 end;
47
48 procedure change(var a:ji; x,y:longint);
49 var
50 mid:longint;
51 begin
52 if (x=a^.l)and(x=a^.r) then
53 begin
54 a^.min:=y;
55 exit;
56 end;
57 mid:=(a^.l+a^.r)>>1;
58 if mid>=x then change(a^.lson,x,y);
59 if mid<x then change(a^.rson,x,y);
60 a^.min:=minn(a^.lson^.min,a^.rson^.min);
61 end;
62
63 begin
64 readln(n,m);
65 for i:=1 to n do read(b[i]);
66 build(a,1,n);
67 for i:=1 to m do
68 begin
69 readln(j,x,y);
70 if j=1 then write(find(a,x,y),' ')
71 else change(a,x,y);
72 end;
73 end.