poj3264 Balanced Lineup 2011-12-20
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Balanced Lineup
Time Limit: 5000MSMemory Limit: 65536K
Total Submissions: 20569Accepted: 9552
Case Time Limit: 2000MS
Description
For the daily milking, Farmer John‘s N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.
Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.
Input
Line 1: Two space-separated integers, N and Q.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.
Output
Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.
Sample Input
6 3
1
7
3
4
2
5
1 5
4 6
2 2
Sample Output
6
3
0
Source
USACO 2007 January Silver
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给定一串数,多次询问某段区间中最大的数减去最小的数
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RMQ,用线段树做的话也是水题。
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1 Program Stone; 2 3 var i,j,k,n,m,lc,rc,best,sest:longint; 4 5 a,max,min:array[1..1 shl 17]of longint; 6 7 function mmax(a,b:longint):longint; 8 9 begin 10 11 if a>b then mmax:=a else mmax:=b; 12 13 end; 14 15 function mmin(a,b:longint):longint; 16 17 begin 18 19 if a>b then mmin:=b else mmin:=a; 20 21 end; 22 23 24 25 procedure built(head,tail,num:longint); //建树 26 27 var i,j,k:longint; 28 29 begin 30 31 if head=tail then begin 32 33 max[num]:=a[head];min[num]:=a[head]; 34 35 exit; 36 37 end; 38 39 k:=(head+tail)div 2; 40 41 built(head,k,num*2); 42 43 built(k+1,tail,num*2+1); 44 45 max[num]:=mmax(max[num*2],max[num*2+1]); //存最大值 46 47 min[num]:=mmin(min[num*2],min[num*2+1]); //存最小值 48 49 end; 50 51 52 53 procedure query(head,tail,num:longint); //询问,线段树操作 54 55 var i,j,k:longint; 56 57 begin 58 59 if (lc<=head)and(tail<=rc) then begin 60 61 best:=mmax(best,max[num]); 62 63 sest:=mmin(sest,min[num]); 64 65 exit; 66 67 end; 68 69 k:=(head+tail)div 2; 70 71 if rc<=k then query(head,k,num*2) else 72 73 if lc>k then query(k+1,tail,num*2+1) else 74 75 begin 76 77 query(head,k,num*2); 78 79 query(k+1,tail,num*2+1); 80 81 end; 82 83 end; 84 85 Begin 86 87 readln(n,m); 88 89 for i:=1 to n do readln(a[i]); 90 91 built(1,n,1); 92 93 for i:=1 to m do 94 95 begin 96 97 readln(lc,rc); 98 99 sest:=1000000;best:=0; 100 101 query(1,n,1); 102 103 writeln(best-sest); 104 105 end; 106 107 end. 108 109
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