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.
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
分析：线段树求最大值和最小值，然后最大值减去最小值即为正解！貌似这题好像有暴力写法？
下面给出AC代码：
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)// 构建线段树 17 { 18 int mid; 19 tree[root].left=left; 20 tree[root].right=right;// 当前节点所表示的区间 21 if(left==right)// 左右区间相同，则此节点为叶子，max 应储存对应某个学生的值 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;// 递归建立左右子树，并从子树中获得最大值 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)// 从节点 root 开始，查找 left 和 right 之间的最大值 35 { 36 int mid; 37 //if(tree[root].left>right||tree[root].right<left)// 若此区间与 root 所管理的区间无交集 38 //return; 39 if(left==tree[root].left&&tree[root].right==right)// 若此区间包含 root 所管理的区间 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 ",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 ",Max-Min); 86 } 87 } 88 return 0; 89 }