【集训Day1】 Dwarves line up

Dwarves line up

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解题思路

用线段树维护每个人当前的位置的区间最大值和最小值，查询时若区间的长度和位置的最大值与最小值的差相对，则已经形成。

code

#include<iostream>
#include<cstdio>
using namespace std;

int n,m;
int a[200010];

struct abc{
    
	int maxn,minn;
}tree[800010];

abc mix(abc a,abc b)
{
    
	return (abc){
    max(a.maxn,b.maxn),min(a.minn,b.minn)};
}

void in(int num,int l,int r,int x,int a)
{
    
	tree[num].maxn=max(tree[num].maxn,a);
	tree[num].minn=min(tree[num].minn,a);
	if(l==r)
		return;
	int mid=(l+r)/2;
	if(x<=mid)
		in(num*2,l,mid,x,a);
	else
		in(num*2+1,mid+1,r,x,a);
}

void chang(int num,int l,int r,int x,int a)
{
    
	if(l==r)
	{
    
		tree[num]=(abc){
    a,a};
		return;
	}
	int mid=(l+r)/2;
	if(x<=mid)
		chang(num*2,l,mid,x,a);
	else
		chang(num*2+1,mid+1,r,x,a);
	tree[num].maxn=max(tree[num*2].maxn,tree[num*2+1].maxn);
	tree[num].minn=min(tree[num*2].minn,tree[num*2+1].minn);
}

abc fd(int num,int l,int r,int a,int b)
{
    
	if(l==a&&r==b)
		return tree[num];
	int mid=(l+r)/2;
	if(b<=mid)
		return fd(num*2,l,mid,a,b);
	else if(a>mid)
		return fd(num*2+1,mid+1,r,a,b);
	else
		return mix(fd(num*2,l,mid,a,mid),fd(num*2+1,mid+1,r,mid+1,b));
}

int main()
{
    
	cin>>n>>m;
	for(int i=1;i<=n*4;i++) tree[i].minn=0x3f3f3f3f;
	for(int i=1;i<=n;i++)
	{
    
		scanf("%d",&a[i]);
		in(1,1,n,a[i],i);
	}
	while(m--)
	{
    
		int	x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		if(x==1)
		{
    
			chang(1,1,n,a[z],y);
			chang(1,1,n,a[y],z);
			swap(a[z],a[y]);
		}
		else
		{
    
			abc t=fd(1,1,n,y,z);
			if(t.maxn-t.minn==z-y)
				printf("YES\n");
			else
				printf("NO\n");
		}
	}
}