pku3264 Balanced Lineup
求区间最值之差，需要Qlogn的算法，可以用线段树或者ST。
线段树1750ms
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1 program pku3264_line_tree(input,output); 2 type 3 node = record 4 left,right,x,y,minx,maxx : longint; 5 end; 6 var 7 tree : array[0..500000] of node; 8 n,q : longint; 9 a : array[0..200000] of longint; 10 tot : longint; 11 function min(aa,bb :longint ):longint; 12 begin 13 if aa<bb then 14 exit(aa); 15 exit(bb); 16 end; { min } 17 function max(aa,bb :longint ):longint; 18 begin 19 if aa>bb then 20 exit(aa); 21 exit(bb); 22 end; { max } 23 procedure build(xx,yy: longint ); 24 var 25 mid,now : longint; 26 begin 27 inc(tot); 28 tree[tot].x:=xx; 29 tree[tot].y:=yy; 30 if xx=yy then 31 begin 32 tree[tot].minx:=a[xx]; 33 tree[tot].maxx:=a[xx]; 34 exit; 35 end; 36 now:=tot; 37 mid:=(xx+yy)>>1; 38 tree[now].left:=tot+1; 39 build(xx,mid); 40 tree[now].right:=tot+1; 41 build(mid+1,yy); 42 tree[now].minx:=min(tree[tree[now].left].minx,tree[tree[now].right].minx); 43 tree[now].maxx:=max(tree[tree[now].left].maxx,tree[tree[now].right].maxx); 44 end; { build } 45 procedure init; 46 var 47 i : longint; 48 begin 49 readln(n,q); 50 for i:=1 to n do 51 readln(a[i]); 52 build(1,n); 53 end; { init } 54 function getmax(now,xx,yy :longint ):longint; 55 var 56 mid : longint; 57 begin 58 if (tree[now].x=xx)and(tree[now].y=yy) then 59 exit(tree[now].maxx); 60 mid:=(tree[now].x+tree[now].y)>>1; 61 if yy<=mid then 62 exit(getmax(tree[now].left,xx,yy)) 63 else 64 if xx>mid then 65 exit(getmax(tree[now].right,xx,yy)) 66 else 67 exit(max(getmax(tree[now].left,xx,mid),getmax(tree[now].right,mid+1,yy))); 68 end; { getmax } 69 function getmin(now,xx,yy :longint ):longint; 70 var 71 mid : longint; 72 begin 73 if (tree[now].x=xx)and(tree[now].y=yy) then 74 exit(tree[now].minx); 75 mid:=(tree[now].x+tree[now].y)>>1; 76 if yy<=mid then 77 exit(getmin(tree[now].left,xx,yy)) 78 else 79 if xx>mid then 80 exit(getmin(tree[now].right,xx,yy)) 81 else 82 exit(min(getmin(tree[now].left,xx,mid),getmin(tree[now].right,mid+1,yy))); 83 end; { getmin } 84 procedure solve(); 85 var 86 i,xx,yy : longint; 87 begin 88 for i:=1 to q do 89 begin 90 readln(xx,yy); 91 writeln(getmax(1,xx,yy)-getmin(1,xx,yy)); 92 end; 93 end; { solve } 94 begin 95 init; 96 solve; 97 end.
ST算法1204ms
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1 program pku3264(input,output); 2 var 3 f,g : array[0..200001,0..18] of longint; 4 n,q : longint; 5 procedure init; 6 var 7 i : longint; 8 begin 9 readln(n,q); 10 for i:=1 to n do 11 begin 12 readln(f[i,0]); 13 g[i,0]:=f[i,0]; 14 end; 15 end; { init } 16 function min(aa,bb :longint ):longint; 17 begin 18 if aa<bb then 19 exit(aa); 20 exit(bb); 21 end; { min } 22 function max(aa,bb :longint ):longint; 23 begin 24 if aa>bb then 25 exit(aa); 26 exit(bb); 27 end; { max } 28 procedure main; 29 var 30 i,j : longint; 31 begin 32 for j:=1 to trunc(ln(n)/ln(2)) do 33 for i:=1 to n do 34 if i+1<<(j-1)>n then 35 break 36 else 37 begin 38 f[i,j]:=min(f[i,j-1],f[i+1<<(j-1),j-1]); 39 g[i,j]:=max(g[i,j-1],g[i+1<<(j-1),j-1]); 40 end; 41 end; { main } 42 procedure solve(); 43 var 44 i,x,y,t : longint; 45 begin 46 for i:=1 to q do 47 begin 48 readln(x,y); 49 t:=trunc(ln(y-x+1)/ln(2)); 50 writeln(max(g[x,t],g[y-1<<t+1,t])-min(f[x,t],f[y-1<<t+1,t])); 51 end; 52 end; { solve } 53 begin 54 init; 55 main; 56 solve; 57 end.
还是ST简洁啊，DP就是NB！
