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| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 42349 | Accepted: 19917 | |
| 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
Output
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
线段树维护区间最大值、最小值，然后相减。AC代码：
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <map> 5 #include <string> 6 #include <string.h> 7 #include <queue> 8 #include <vector> 9 #include <set> 10 #include <cmath> 11 #define inf 0x7fffffff 12 using namespace std; 13 const int maxn=50000; 14 struct b{ 15 int imax,imin; 16 }tree[maxn*4+100]; 17 int n,q,x,y,a[maxn+100]; 18 void build(int p,int l,int r){ 19 if(l==r) {tree[p].imax=a[l],tree[p].imin=a[l];return ;} 20 int mid=(l+r)/2; 21 build(p*2,l,mid); 22 build(p*2+1,mid+1,r); 23 tree[p].imax=max(tree[p*2].imax,tree[p*2+1].imax); 24 tree[p].imin=min(tree[p*2].imin,tree[p*2+1].imin); 25 } 26 int findmin(int p,int l,int r,int x,int y){ 27 if(x<=l&&r<=y) return tree[p].imin; 28 int mid=(l+r)/2; 29 if(x>mid) return findmin(p*2+1,mid+1,r,x,y); 30 if(y<=mid) return findmin(p*2,l,mid,x,y); 31 return min(findmin(p*2+1,mid+1,r,x,y),findmin(p*2,l,mid,x,y)); 32 } 33 int findmax(int p,int l,int r,int x,int y){ 34 if(x<=l&&r<=y) return tree[p].imax; 35 int mid=(l+r)/2; 36 if(x>mid) return findmax(p*2+1,mid+1,r,x,y); 37 if(y<=mid) return findmax(p*2,l,mid,x,y); 38 return max(findmax(p*2+1,mid+1,r,x,y),findmax(p*2,l,mid,x,y)); 39 } 40 int main() 41 { 42 scanf("%d%d",&n,&q); 43 for(int i=1;i<=n;i++) 44 scanf("%d",&a[i]); 45 build(1,1,n); 46 for(int i=1;i<=q;i++){ 47 scanf("%d%d",&x,&y); 48 printf("%d\n",findmax(1,1,n,x,y)-findmin(1,1,n,x,y)); 49 } 50 return 0; 51 }
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原文地址：http://www.cnblogs.com/GeniusYang/p/5225783.html
