POJ3264 Balanced Lineup(线段树,区间最值)
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题目:
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
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
Sample Input
6 31734251 54 62 2
Sample Output
630
Source
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思路:简单的区间最值问题,记录最大值的时候同时记录一下最小值,询问的时候开两个函数,分别询问最大值和最小值。
代码:
#include <cstdio>#include <cstring>#include <cctype>#include <string>#include <set>#include <iostream>#include <stack>#include <cmath>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define N 60050#define ll long longusing namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1int MAX[N<<2],MIN[N<<2];void pushup(int rt){MAX[rt]=max(MAX[rt<<1],MAX[rt<<1|1]);MIN[rt]=min(MIN[rt<<1],MIN[rt<<1|1]);}void build(int l,int r,int rt){if(l==r){scanf("%d",&MAX[rt]);MIN[rt]=MAX[rt];return;}int m=(l+r)>>1;build(lson);build(rson);pushup(rt);}int query1(int L,int R,int l,int r,int rt){if(L<=l&&r<=R){return MAX[rt];}int m=(l+r)>>1;int ret=0;if(L<=m) ret=max(ret,query1(L,R,lson));if(m<R) ret=max(ret,query1(L,R,rson));return ret;}int query2(int L,int R,int l,int r,int rt){if(L<=l&&r<=R){return MIN[rt];}int m=(l+r)>>1;int ret=inf;if(L<=m) ret=min(ret,query2(L,R,lson));if(m<R) ret=min(ret,query2(L,R,rson));return ret;}int main(){int n,q,a,b;while(~scanf("%d%d",&n,&q)){mem(MAX,0);build(1,n,1);while(q--){scanf("%d%d",&a,&b);printf("%d\n",query1(a,b,1,n,1)-query2(a,b,1,n,1));}}return 0;}
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