poj 3264 线段树 寻找最大最小值 SEGMENT TREE

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

My original code is lengthy:

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int N,Q,cows[50001],A,B;
int cal(int a,int b){
    int max=cows[a],min=cows[a];
    for(int i=a+1;i<=b;i++){
        if(max<cows[i])
           max=cows[i];
        if(min>cows[i])
           min=cows[i];
    }
    return max-min;
}
struct node{
    int l,r,min,max;   
}tree[50001*3];
struct maxmin{
    int max,min;
};

void build(int k,int l,int r){
    
    tree[k].l=l;
    tree[k].r=r;    
    if( l==r ){
        tree[k].max=max(cows[l],cows[r]);
        tree[k].min=min(cows[l],cows[r]);
    }
    else{
        build(2*k+1,l,(l+r)/2);
        build(2*k+2,(l+r)/2+1,r);
        tree[k].max=max(tree[2*k+1].max,tree[2*k+2].max);
        tree[k].min=min(tree[2*k+1].min,tree[2*k+2].min);                
    }
}
int query(int k,int l,int r,int *Max,int *Min){
    int max1,max2,min1,min2;
    
    if(tree[k].l==l && tree[k].r==r){
        (*Max)=tree[k].max;
        (*Min)=tree[k].min;
        return (*Max)-(*Min);
    }
    
    int mid=(tree[k].l+tree[k].r)/2;
    if( r<=mid ){
        query(2*k+1,l,r,Max,Min);
        return (*Max)-(*Min);
    }
    else if( l>=mid+1 ){
        query(2*k+2,l,r,Max,Min);
        return (*Max)-(*Min);
    }
    else{
        query(2*k+1,l,mid,Max,Min);
        max1=(*Max);
        min1=(*Min);
        query(2*k+2,mid+1,r,Max,Min);
        max2=(*Max);
        min2=(*Min);
        (*Max)=max(max1,max2);
        (*Min)=min(min1,min2);
        
        return (*Max)-(*Min);        
    }
    
}
int main(){
    int i,j,Max,Min;
    scanf("%d%d",&N,&Q);
    for(i=1;i<=N;i++)
       scanf("%d",&cows[i]);
    build(0,1,N);
    
    for(i=0;i<Q;i++){
       scanf("%d%d",&A,&B);
       int res=query( 0,A,B,&Max,&Min );
       printf("%d\n",res);
    }
    //system("pause");
    return 0; 
}

The concise code :

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int N,Q,cows[50001],A,B;

struct node{
  int l, r, min, max;   
}tree[50001 * 3];
int maxx, minx;

void build(int k, int l, int r){
    
  tree[k].l=l;
  tree[k].r=r;    
  if( l==r )
    tree[k].max=tree[k].min=cows[l];
  else{
    build(2 * k + 1, l, (l + r) / 2);
    build(2 * k + 2, (l + r) / 2 + 1, r);
    tree[k].max = max(tree[2*k+1].max, tree[2*k+2].max);
    tree[k].min = min(tree[2*k+1].min, tree[2*k+2].min);                
  }
}
int query(int k, int l, int r){
  int max1, max2, min1, min2;    
  if(tree[k].l == l && tree[k].r == r){
    maxx = tree[k].max;
    minx = tree[k].min;
    return (maxx - minx);
  }    
  int mid = (tree[k].l + tree[k].r) / 2;
  if(r <= mid){
    query(2 * k + 1, l, r);
    return (maxx - minx);
  }
  else if(l >= mid + 1){
    query(2 * k + 2, l, r);
    return (maxx - minx);
  }
  else{
    query(2 * k + 1, l, mid);
    max1 = maxx;
    min1 = minx;
    query(2 * k + 2, mid + 1, r);    
    maxx = max(max1, maxx);
    minx = min(min1, minx);
    return maxx - minx;        
  }
    
}
int main(){
  int i,j,Max,Min;
  scanf("%d%d",&N,&Q);
  for(i=1;i<=N;i++)
    scanf("%d",&cows[i]);
  build(0,1,N);
    
  for(i=0;i<Q;i++){
    scanf("%d%d",&A,&B);
    int res=query(0,A,B);
    printf("%d\n",res);
  }
  //system("pause");
  return 0; 
}

Another way:

Use the RMQ to get the maximum and minimum of d[i][j]:

//8200K    1610MS
#include
#include
const int M = 50005;
const int N = 20;
int H,dx[M][N],dy[M][N],n;
int max (int a,int b)
{
    return a > b ? a : b;
}
int min(int a,int b)
{
    return a > b ? b : a;
}
void Init()
{
    int i,j;
    for (i = 1; i <= n; i ++)
    {
        scanf ("%d",&H);
        dx[i][0] = dy[i][0] = H;
    }
    for (j = 1; (1<<j) <= n; j ++)
        for (i = 1; i +(1<<j)-1<=n; i ++)
        {
            dx[i][j] = min(dx[i][j-1],dx[i+(1<<(j-1))][j-1]);
            dy[i][j] = max(dy[i][j-1],dy[i+(1<<(j-1))][j-1]);
        }

}
int RMQ(int L ,int R)
{
    int k = 0;
    while ((1<<(k+1))<=R-L+1) k ++;
    return max(dy[L][k],dy[R-(1<<k)+1][k]) - min(dx[L][k],dx[R-(1<<k)+1][k]);
}
int main ()
{
    int m,a,b;
    while (~scanf ("%d%d",&n,&m))
    {
        Init();
        while (m --)
        {
            scanf ("%d%d",&a,&b);
            printf ("%d\n",RMQ(a,b));
        }
    }
    return 0;
}

A better concise version:

#include <stdio.h>
#include <vector>
#include <iostream>
using namespace std;

class Node {
 public:
  int l, r, max, min;
  Node(int ll = 0, int rr = 0, int max1 = 0, int min1 = 0) : l(ll), r(rr), max(max1), min(min1){}
};

void build(vector<Node>& tree, const vector<int>& height, int k, int l, int r) {
  tree[k].l = l, tree[k].r = r;
  if (l == r) {
    tree[k].max = tree[k].min = height[l];
    return;
  }
  build(tree, height, k*2+1, l, (l+r)/2);
  build(tree, height, k*2+2, (l+r)/2+1, r);
  tree[k].max = max(tree[k*2+1].max, tree[k*2+2].max);
  tree[k].min = min(tree[k*2+1].min, tree[k*2+2].min); 
}
int query(vector<Node>& tree, int k, int l, int r, int& maxv, int& minv) {
  int max1, min1, max2, min2;
  
  if (l == tree[k].l && r == tree[k].r) {
    maxv = tree[k].max, minv = tree[k].min;
    return maxv - minv;
  }
  int mid = (tree[k].l + tree[k].r) / 2;
  if (r <= mid) {
    return query(tree, k*2+1, l, r, maxv, minv);
  }
  else if (l >= mid+1) {
    return query(tree, k*2+2, l, r, maxv, minv);
  }
  else {
    query(tree, k*2+1, l, mid, max1, min1);
    query(tree, k*2+2, mid+1, r, max2, min2);   
    maxv = max(max1, max2);
    minv = min(min1, min2);
    return maxv -minv;
  }  
}
int main() {
  int N, Q, l, r, maxv, minv;

  scanf("%d%d", &N, &Q);
  vector<int> height(N,0);
  vector<Node> tree(50001 * 3);
  
  for (int i = 0; i < N; ++i)
    scanf("%d", &height[i]);
  build(tree, height, 0, 0, N-1);
  for (int i = 0; i < Q; ++i) {
    scanf("%d%d", &l, &r);
    printf("%d\n", query(tree, 0, l-1, r-1, maxv, minv));
  }
  
  return 0;
}