BZOJ 1636: [Usaco2007 Jan]Balanced Lineup
题目
1636: [Usaco2007 Jan]Balanced Lineup
Time Limit: 5 Sec Memory Limit: 64 MBDescription
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
Max difference in height
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
* 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.
to a reply and indicates the difference in height between the
tallest and shortest cow in the range.
Sample Output
6
3
0
HINT
Source
题解
静态线段树水题Orz，区间查找最大值和最小值做差就可以了Orz
代码
1 /*Author:WNJXYK*/ 2 #include<cstdio> 3 using namespace std; 4 5 struct Btree{ 6 int left,right; 7 int num; 8 int max,min; 9 }; 10 const int Maxn=50000; 11 Btree tree[Maxn*4+10]; 12 13 int n,q; 14 int num[Maxn+10]; 15 16 inline int remin(int a,int b){ 17 if (a<b) return a; 18 return b; 19 } 20 21 inline int remax(int a,int b){ 22 if (a>b) return a; 23 return b; 24 } 25 26 void build(int x,int left,int right){ 27 tree[x].left=left; 28 tree[x].right=right; 29 if (left==right){ 30 tree[x].max=tree[x].min=tree[x].num=num[left]; 31 }else{ 32 int mid=(left+right)/2; 33 build(x*2,left,mid); 34 build(x*2+1,mid+1,right); 35 tree[x].max=remax(tree[x*2].max,tree[x*2+1].max); 36 tree[x].min=remin(tree[x*2].min,tree[x*2+1].min); 37 } 38 } 39 40 int queryMax(int x,int left,int right){ 41 if (left<=tree[x].left && tree[x].right<=right){ 42 return tree[x].max; 43 }else{ 44 int res=0; 45 int mid=(tree[x].left+tree[x].right)/2; 46 if (left<=mid) res=remax(res,queryMax(x*2,left,right)); 47 if (mid+1<=right) res=remax(res,queryMax(x*2+1,left,right)); 48 return res; 49 } 50 } 51 52 int queryMin(int x,int left,int right){ 53 if (left<=tree[x].left && tree[x].right<=right){ 54 return tree[x].min; 55 }else{ 56 int res=2147483647; 57 int mid=(tree[x].left+tree[x].right)/2; 58 if (left<=mid) res=remin(res,queryMin(x*2,left,right)); 59 if (mid+1<=right) res=remin(res,queryMin(x*2+1,left,right)); 60 return res; 61 } 62 } 63 64 int main(){ 65 scanf("%d%d",&n,&q); 66 for (int i=1;i<=n;i++) scanf("%d",&num[i]); 67 build(1,1,n); 68 for (int i=1;i<=q;i++){ 69 int left,right; 70 scanf("%d%d",&left,&right); 71 printf("%d\n",queryMax(1,left,right)-queryMin(1,left,right)); 72 } 73 return 0; 74 }
