POJ 3264 Balanced Lineup

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 53629		Accepted: 25223
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

典型RMQ问题，目前只会写数组类线段树，但这道题需要存最大最小值，结构体类型线段树不是很熟，所以果断写了RMQ

时间：3694ms

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
#define N 50006
int maxx[N][25],minx[N][25];
int n,m,x,y;
void get_RMQ()
{
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&x);
        maxx[i][0]=x;
        minx[i][0]=x;
    }
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=1;i+(1<<j)-1<=n;i++)
        {
            maxx[i][j]=max(maxx[i][j-1],maxx[i+(1<<(j-1))][j-1]);//存最大值
            minx[i][j]=min(minx[i][j-1],minx[i+(1<<(j-1))][j-1]);//存最小值
        }
    }
}
int get_K(int l,int r)
{
    if(l>r) return 0;
    int k=0;
    while(1<<(1+k)<=r-l+1) k++;
    return k;//查找区间即j
}
int RMQUPPER(int l,int r,int k)
{
    return max(maxx[l][k],maxx[r-(1<<k)+1][k]);
}
int RMQLOWER(int l,int r,int k)
{
    return min(minx[l][k],minx[r-(1<<k)+1][k]);
}
int main()
{
    scanf("%d%d",&n,&m);
    get_RMQ();
    while(m--)
    {
        int k;
        scanf("%d%d",&x,&y);
        k=get_K(x,y);
        printf("%d\n",RMQUPPER(x,y,k)-RMQLOWER(x,y,k));
    }
}

再来一个线段树代码，注意必须两次查询，一次查最大，一次查最小，我想把最大最小值相减差值保存在结构体中，只查询一次，但这样会wrong answer

见代码注释部分

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
#define N 50009
#define mod 0x3f3f3f3f
int n,m,x,y;
struct Node
{
  int mx;
  int mn;
  //int mxn;
}tree[N*4];
void buildtree(int l,int r,int node)
{
  int mid=(l+r)>>1;
  if(l==r)
  {
    scanf("%d",&tree[node].mx);
    tree[node].mn=tree[node].mx;
    //tree[node].mxn=tree[node].mx-tree[node].mn;
    return;
  }
  buildtree(l,mid,node<<1);
  buildtree(mid+1,r,(node<<1)+1);
  tree[node].mx=max(tree[node<<1].mx,tree[(node<<1)+1].mx);
  tree[node].mn=min(tree[node<<1].mn,tree[(node<<1)+1].mn);
  //tree[node].mxn=tree[node].mx-tree[node].mn;
}
int queryu(int l,int r,int ll,int rr,int node)
{
  int k=0;
  int mid=(ll+rr)>>1;
  if(l<=ll && r>=rr) return tree[node].mx;
  if(l<=mid) k=max(k,queryu(l,r,ll,mid,node<<1));
  if(r>mid) k=max(k,queryu(l,r,mid+1,rr,(node<<1)+1));
  return k;
}
int queryt(int l,int r,int ll,int rr,int node)
{
  int k=mod;
  int mid=(ll+rr)>>1;
  if(l<=ll && r>=rr) return tree[node].mn;
  if(l<=mid) k=min(k,queryt(l,r,ll,mid,node<<1));
  if(r>mid) k=min(k,queryt(l,r,mid+1,rr,(node<<1)+1));
  return k;
}
int main()
{
  scanf("%d%d",&n,&m);
  buildtree(1,n,1);
  while(m--)
  {
    scanf("%d%d",&x,&y);
    printf("%d\n",queryu(x,y,1,n,1)-queryt(x,y,1,n,1));
  }
  return 0;
}