poj_3264 Balanced Lineup
Balanced LineupTime Limit: 5000MS
Memory Limit: 65536K
Total Submissions: 23380
Accepted: 10882
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 agame of Ultimate Frisbee with some of the cows. To keep things 易做图, he willtake 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 inheight.
Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potentialgroups of cows and their heights (1 ≤ height ≤ 1,000,000). For eachgroup, he wants your help to determine the difference in height between theshortest 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 theheight 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 Binclusive.
Output
Lines 1..Q: Each line contains a single integerthat is a response to a reply and indicates the difference in height betweenthe 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 JanuarySilver
解题思路：
这个题目可以用线段树来做，只有两个操作，第一个建立树，第二个查询线段区间即可。
代码：
#include <iostream>
#include<cstdio>
#include<cstring>
#define N 50005
#define Q 200005
using namespace std;
struct segment
{
int left;
int right;
int max;
int min;
int value;
};
int n,q;
int high;
segment seg[N*4];
int max(int a,int b)
{
if(a>b)
return a;
return b;
}
int min(int a,int b)
{
if(a<b)
return a;
return b;
}
void update(int rt)
{
seg[rt].max=max(seg[rt*2].max,seg[rt*2+1].max);
seg[rt].min=min(seg[rt*2].min,seg[rt*2+1].min);
}
//建树
void buildTree(int l,int r,int rt)
{
seg[rt].left=l;
seg[rt].right=r;
if(l==r)
{
scanf("%d",&high);
seg[rt].max=seg[rt].min=high;
return;
}
int mid=(l+r)/2;
buildTree(l,mid,rt*2);
buildTree(mid+1,r,rt*2+1);
update(rt);
}
//查询区间a到b中的最大差值
int queryMax(int a,int b,int rt)
{
if(a==seg[rt].left && b==seg[rt].right)
{
return seg[rt].max;
}
int res=0;
int mid=(seg[rt].left+seg[rt].right)/2;
if(b<=mid)
{
res=max(res,queryMax(a,b,rt*2));
}
else if(a>mid)
{
res=max(res,queryMax(a,b,rt*2+1));
}
else
{
res=max(res,queryMax(a,mid,rt*2));
res=max(res,queryMax(mid+1,b,rt*2+1));
}
return res;
}
int queryMin(int a,int b,int rt)
{
if(a==seg[rt].left && b==seg[rt].right)
{
return seg[rt].min;
}
int mid=(seg[rt].left+seg[rt].right)/2;
int res=999999999;
if(b<=mid)
{
res=min(res,queryMin(a,b,rt*2));
}
else if(a>mid)
{
res=min(res,queryMin(a,b,rt*2+1));
}
else
{
res=min(res,queryMin(a,mid,rt*2));
res=min(res,queryMin(mid+1,b,rt*2+1));
}
return res;
}
int main()
{
int a,b;
scanf("%d%d",&n,&q);
buildTree(1,n,1);
while(q--)
{
scanf("%d%d",&a,&b);
int max = queryMax(a,b,1);
int min= queryMin(a,b,1);
int val=max-min;
printf("%d\n",val);
}
return 0;
}
补充：软件开发 , C++ ,
