[POJ] 3264 Balanced Lineup [线段树]

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 34306		Accepted: 16137
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

Sample Output

6
3
0

Source

USACO 2007 January Silver

题解：典型的RMQ问题。线段树的应用。

代码：

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<math.h>
  4 #include<ctype.h>
  5 #include<stdlib.h>
  6 #include<stdbool.h>
  7 
  8 #define rep(i,a,b)      for(i=(a);i<=(b);i++)
  9 #define clr(x,y)        memset(x,y,sizeof(x))
 10 #define sqr(x)          (x*x)
 11 #define LL              long long
 12 
 13 const int INF=0xffffff0;
 14    
 15 struct {
 16     int L,R;
 17     int minV,maxV;
 18 } tree[800010];
 19     
 20 int i,j,n,q,minV,maxV;
 21 
 22 int min(int a, int b)
 23 {
 24     if(a<b) return a;
 25     return b;
 26 }
 27 
 28 int max(int a,int b)
 29 {
 30     if(a>b) return a;
 31     return b;
 32 }
 33 
 34 void BuildTree(int root,int L,int R)
 35 {
 36     tree[root].L=L;
 37     tree[root].R=R;
 38     tree[root].maxV=-INF;
 39     tree[root].minV=INF;
 40     
 41     if(L!=R) {
 42         BuildTree(2*root,L,(L+R)/2);
 43         BuildTree(2*root+1,(L+R)/2+1,R);
 44     }
 45     
 46 }
 47 
 48 void Insert(int root,int i,int v)
 49 {
 50     int mid;
 51 
 52     if(tree[root].L==tree[root].R) {
 53         tree[root].maxV=tree[root].minV=v;
 54         return ;
 55     }
 56     
 57     tree[root].minV=min(tree[root].minV,v);
 58     tree[root].maxV=max(tree[root].maxV,v);
 59     
 60     mid=(tree[root].L+tree[root].R)/2;
 61     if(i<=mid) 
 62         Insert(2*root,i,v);
 63     else
 64         Insert(2*root+1,i,v);
 65     
 66 }
 67 
 68 void Query(int root,int s,int e)
 69 {
 70     int mid;
 71     
 72     if(tree[root].minV>=minV && tree[root].maxV<=maxV) return ;
 73     if(tree[root].L==s  &&  tree[root].R==e) {
 74         minV=min(minV,tree[root].minV);
 75         maxV=max(maxV,tree[root].maxV);
 76         return ;
 77     }
 78     
 79     mid=(tree[root].L+tree[root].R)/2;
 80     
 81     if(e<=mid) 
 82         Query(2*root,s,e);
 83     else if(s>mid)
 84         Query(2*root+1,s,e);
 85     else {
 86             Query(2*root,s,mid);
 87             Query(2*root+1,mid+1,e);
 88     } 
 89     
 90 }
 91 
 92 
 93 int main()
 94 {
 95     int i,x,y;
 96     
 97     scanf("%d%d",&n,&q);
 98     BuildTree(1,1,n);
 99     rep(i,1,n) {
100         scanf("%d",&x);
101         Insert(1,i,x);
102     }
103     
104     while(q--) {
105         scanf("%d%d",&x,&y);
106         minV=INF;
107         maxV=-INF;
108         Query(1,x,y);
109         printf("%d
",maxV-minV);
110     }
111     
112     return 0;
113 }
114 
115

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原文地址：https://www.cnblogs.com/sxiszero/p/3916558.html