Balanced Lineup(线段树)

发布于 2020-10-08  0 次阅读


题目大意:

Balanced Lineup
一片绿地的N(1≤N≤50000)颗树排成一排,q次查询,每次查询区间最大值和最小值之差

思路:

很ez的线段树了,更新和push_down都不需要,lazy标记也无
直接线段树维护区间最大值和最小值即可

AC Code:

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
typedef pair<int, int > PII;
#define debug(a) cout<<#a<<"="<<a<<endl;
#define INF 0x3f3f3f3f
const int N=5e4 +9;
int n, q, x, y;
int a[N];
struct segtree{
    int l, r;
	int minv, maxv;
}tr[N<<2];
inline int read()
{
    int ans=0;
    char last=' ',ch=getchar();
    while(ch<'0'||ch>'9') last=ch,ch=getchar();
    while(ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
    if(last=='-') ans=-ans;
    return ans;
}
inline int lc(int p) {return p<<1;}
inline int rc(int p) {return p<<1|1;}
inline void push_up(int p){
    tr[p].maxv=max(tr[lc(p)].maxv, tr[rc(p)].maxv);
    tr[p].minv=min(tr[lc(p)].minv, tr[rc(p)].minv);
}
inline void build(int p, int l, int r){
    tr[p].l=l, tr[p].r=r;
	if(l==r)	{tr[p].maxv=tr[p].minv=a[l]; return ;}
	int mid=(l+r)>>1;
	build(lc(p), l, mid);
	build(rc(p), mid+1, r);
    push_up(p);
}
inline PII query(int p, int l, int r,int x, int y){
    if(x>r || y<l)	return PII(-INF, INF);
	if(l<=x && y<=r) return PII(tr[p].maxv, tr[p].minv);
    int mid=(x+y)>>1;
    PII ql=query(lc(p), l, r, x, mid);
    PII qr=query(rc(p), l, r, mid+1, y);
    return PII(max(ql.first, qr.first), min(ql.second, qr.second));
}
int main(){
    n=read(); q=read();
    for(int i=1; i<=n; i++) a[i]=read();
    build(1,1,n);
    for(int i=1; i<=q; i++){
        x=read(); y=read();
        PII a=query(1,x,y,1,n);
        cout<<a.first-a.second<<endl;
    }
	return 0;   
}

平平无奇的在校大学生