Balanced Lineup

Time Limit: 5000MS	?	Memory Limit: 65536K
Total Submissions: 53703	?	Accepted: 25237
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 題目鏈接：http://poj.org/problem?id=3264 分析：線段樹求最大值和最小值，然后最大值減去最小值即為正解！貌似這題好像有暴力寫法？ 下面給出AC代碼： 1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 #define maxsize 200020 6 typedef struct 7 { 8 int left,right; 9 int maxn; 10 int minn; 11 }Node; 12 int n,m; 13 int Max,Min; 14 int num[maxsize]; 15 Node tree[maxsize*20]; 16 inline void buildtree(int root,int left,int right)// 構建線段樹 17 { 18 int mid; 19 tree[root].left=left; 20 tree[root].right=right;// 當前節點所表示的區間 21 if(left==right)// 左右區間相同，則此節點為葉子，max 應儲存對應某個學生的值 22 { 23 tree[root].maxn=num[left]; 24 tree[root].minn=num[left]; 25 return; 26 } 27 mid=(left+right)/2; 28 //int a,b;// 遞歸建立左右子樹，并從子樹中獲得最大值 29 buildtree(2*root,left,mid); 30 buildtree(2*root+1,mid+1,right); 31 tree[root].maxn=max(tree[root*2].maxn,tree[root*2+1].maxn); 32 tree[root].minn=min(tree[root*2].minn,tree[root*2+1].minn); 33 } 34 inline void find(int root,int left,int right)// 從節點 root 開始，查找 left 和 right 之間的最大值 35 { 36 int mid; 37 //if(tree[root].left>right||tree[root].right<left)// 若此區間與 root 所管理的區間無交集 38 //return; 39 if(left==tree[root].left&&tree[root].right==right)// 若此區間包含 root 所管理的區間 40 { 41 Max=max(tree[root].maxn,Max); 42 Min=min(tree[root].minn,Min); 43 return; 44 } 45 mid=(tree[root].left+tree[root].right)/2; 46 if(right<=mid) 47 find(root*2,left,right); 48 else if(left>mid) 49 find(root*2+1,left,right); 50 else 51 { 52 find(root*2,left,mid); 53 find(root*2+1,mid+1,right); 54 //tree[root].maxn=max(tree[root*2].maxn,tree[root*2+1].maxn); 55 //tree[root].minn=min(tree[root*2].minn,tree[root*2+1].minn); 56 //return; 57 } 58 } 59 60 int main() 61 { 62 //char c; 63 int i; 64 int x,y; 65 //scanf("d%d",&n,&m); 66 while(scanf("%d%d",&n,&m)!=EOF) 67 { 68 for(i=1;i<=n;i++) 69 scanf("%d",&num[i]); 70 buildtree(1,1,n); 71 for(i=1;i<=m;i++) 72 { 73 //getchar(); 74 Max=-99999999999; 75 Min= 99999999999; 76 scanf("%d%d",&x,&y); 77 //if(c=='Q') 78 //printf("%d\n",find(1,x,y)); 79 //else 80 //{ 81 // num[x]=y; 82 // update(1,x,y); 83 //} 84 find(1,x,y); 85 printf("%d\n",Max-Min); 86 } 87 } 88 return 0; 89 }

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轉載于:https://www.cnblogs.com/ECJTUACM-873284962/p/7133096.html

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