【WOJ 4348】ldx的SGT水题1

版权声明:虽然我很菜,不过转载请标明出处。 https://blog.csdn.net/Patrickpwq/article/details/86665282

水题 把排序看成区间覆盖就行 

我写复杂了 只需要维护0的个数就行了的

#include<bits/stdc++.h>
const int N=1000005;
using namespace std;
int n,v[N];
struct Tree
{
	int l,r,cover,sum[2];
}tree[4*N];
inline void pushdown(int now)
{
	int f=tree[now].cover;
	if(!f)return;
	tree[2*now].cover=f; tree[2*now+1].cover=f;
	if(f==1)
	{
		tree[2*now].sum[0]=tree[2*now].r-tree[2*now].l+1;
		tree[2*now].sum[1]=0;
		tree[2*now+1].sum[0]=tree[2*now+1].r-tree[2*now+1].l+1;
		tree[2*now+1].sum[1]=0;
	}
	if(f==2)
	{
		tree[2*now].sum[1]=tree[2*now].r-tree[2*now].l+1;
		tree[2*now].sum[0]=0;
		tree[2*now+1].sum[1]=tree[2*now+1].r-tree[2*now+1].l+1;
		tree[2*now+1].sum[0]=0;
	}
	tree[now].cover=0;
}
inline void pushup(int now)
{
	tree[now].sum[0]=tree[2*now].sum[0]+tree[2*now+1].sum[0];
	tree[now].sum[1]=tree[2*now].sum[1]+tree[2*now+1].sum[1];
}
void build(int now,int l,int r)
{
	tree[now].l=l; tree[now].r=r;
	if(l==r)
	{
		tree[now].sum[v[l]]=1;
		return;            
	}
	int m=(l+r)>>1;
	build(2*now,l,m);
	build(2*now+1,m+1,r);
	pushup(now);
}
void write(int now,int l,int r)
{
	if(l==r)
	{
		if(tree[now].sum[0])cout<<"0";else cout<<"1";
		return;            
	}pushdown(now);
	int m=(l+r)>>1;
	write(2*now,l,m);
	write(2*now+1,m+1,r);
}
inline int query(int now,int l,int r)
{ 
	if(l<=tree[now].l&&tree[now].r<=r)
	{
		if(tree[now].cover==1)	return (tree[now].r-tree[now].l+1);
		else if(tree[now].cover==2)	return 0;
		return tree[now].sum[0];
	}
	pushdown(now);
	int m=(tree[now].l+tree[now].r)>>1,ans=0;
	if(l<=m) ans+=query(2*now,l,r);
	if(r>m)	ans+=query(2*now+1,l,r);
	return ans;
}
inline void cover(int now,int l,int r,int x)
{
	if(l<=tree[now].l&&tree[now].r<=r)
	{
		tree[now].cover=x; 
		if(tree[now].cover==1) tree[now].sum[0]=(tree[now].r-tree[now].l+1),tree[now].sum[1]=0;
		else if(tree[now].cover==2)	tree[now].sum[1]=(tree[now].r-tree[now].l+1),tree[now].sum[0]=0;
		return;
	}
	pushdown(now);
	int m=(tree[now].l+tree[now].r)>>1;
	if(l<=m) cover(2*now,l,r,x);
	if(r>m)	cover(2*now+1,l,r,x);
	pushup(now);
}
char ch[N];
int main()
{
	scanf("%s",ch+1);
	n=strlen(ch+1);
	for(int i=1;i<=n;i++)	v[i]=ch[i]-'0';
	int T;
	build(1,1,n);
	cin>>T;
	while(T--)
	{
		int l,r,x;
		scanf("%d%d%d",&l,&r,&x);
		if(x==0)	//升序
		{
			int sum0=query(1,l,r);
			cover(1,l,l+sum0-1,1); cover(1,l+sum0,r,2);
		}
		if(x==1)
		{
			int sum0=query(1,l,r); int sum1=(r-l+1)-sum0; 
			cover(1,l,l+sum1-1,2);
			cover(1,l+sum1,r,1);
		}
	}
	write(1,1,n);
	//for(int i=1;i<=n;i++) printf("%d",(query(1,i,i)^1));
	return 0;
}

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转载自blog.csdn.net/Patrickpwq/article/details/86665282