385
技術社區[雲棲]
【北大夏令營筆記-線段樹】POJ3264-Balanced Lineup
Balanced LineupTime Limit: 5000MS Memory Limit: 65536K
Total Submissions: 32777 Accepted: 15424
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
題意：求給定Q(1<=Q<=200000)個數a1,a2...aq，多次求任意區間內ai-aj中最大數和最小數的差；
思路：線段樹查詢；
AC代碼：
#include<stdio.h>
#include<string.h>
#define MAX_INF 9999999
#define MIN_INF -9999999
struct Node
{
int l,r;
int max,min;
}a[200005];
int max(int n,int m)
{
return n>m?n:m;
}
int min(int n,int m)
{
return n<m?n:m;
}
void Build(int n,int l,int r)
{
a[n].l=l;
a[n].r=r;
a[n].max=MIN_INF;
a[n].min=MAX_INF;
if(l==r)
return;
Build(2*n,l,(l+r)/2);
Build(2*n+1,(l+r)/2+1,r);
}
void Insert(int n,int v,int num)
{
if(a[n].max<num) a[n].max=num;
if(a[n].min>num) a[n].min=num;
if(a[n].l==a[n].r)
return;
if(v<=(a[n].l+a[n].r)/2)
Insert(n*2,v,num);
else
Insert(n*2+1,v,num);
}
int QMax(int n,int l,int r)
{
if(a[n].l==l&&a[n].r==r)
return a[n].max;
if(r<=(a[n].l+a[n].r)/2)
return QMax(n*2,l,r);
else
if(l>=(a[n].l+a[n].r)/2+1)
return QMax(n*2+1,l,r);
else
{
return max(QMax(n*2,l,(a[n].l+a[n].r)/2),QMax(n*2+1,(a[n].l+a[n].r)/2+1,r));
}
}
int QMin(int n,int l,int r)
{
if(a[n].l==l&&a[n].r==r)
return a[n].min;
if(r<=(a[n].l+a[n].r)/2)
return QMin(n*2,l,r);
else
if(l>=(a[n].l+a[n].r)/2+1)
return QMin(n*2+1,l,r);
else
{
return min(QMin(n*2,l,(a[n].l+a[n].r)/2),QMin(n*2+1,(a[n].l+a[n].r)/2+1,r));
}
}
void QMinus(int n,int l,int r)
{
if(a[n].l==l&&a[n].r==r)
{
printf("%d\n",a[n].max-a[n].min);
}
else
{
if(r<=(a[n].l+a[n].r)/2)
QMinus(n*2,l,r);
else
if(l>=(a[n].l+a[n].r)/2+1)
QMinus(n*2+1,l,r);
else
{
int m1=QMax(n*2,l,(a[n].l+a[n].r)/2);
int m2=QMax(n*2+1,(a[n].l+a[n].r)/2+1,r);
int m3=QMin(n*2,l,(a[n].l+a[n].r)/2);
int m4=QMin(n*2+1,(a[n].l+a[n].r)/2+1,r);
printf("%d\n",max(m1,m2)-min(m3,m4));
}
}
}
int main()
{
int i,j,n,m,x,l,r;
while(scanf("%d %d",&n,&m)!=EOF)
{
Build(1,1,n);
for(i=1;i<=n;i++)
{
scanf("%d",&x);
Insert(1,i,x);
}
for(i=0;i<m;i++)
{
scanf("%d %d",&l,&r);
QMinus(1,l,r);
}
}
return 0;
}
最後更新：2017-04-03 05:39:17