附加一道异或树的题目

Posted 2022-02-18 钟钟终

建树加异或操作模拟，无了，实在是很好用，只是记录一下。
还有一个小坑，必须要把字符串转换为int类型存入

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int maxn=2e5+5;
int a[maxn],n,m;
struct node

    int val,lazy;
sg[maxn<<2];
void pushup(int rt)

    sg[rt].val=sg[rt<<1].val+sg[rt<<1|1].val;

void build(int l,int r,int rt)

    if(l==r)
    
        sg[rt].val=a[l];return;
    
    sg[rt].lazy=0;
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    pushup(rt);

void pushdown(int rt,int ln,int rn)

    if(sg[rt].lazy==1)
    
        sg[rt<<1].lazy^=1;
        sg[rt<<1|1].lazy^=1;
        sg[rt<<1].val=ln-sg[rt<<1].val;
        sg[rt<<1|1].val=rn-sg[rt<<1|1].val;
        sg[rt].lazy=0;
    

void update(int L,int R,int l,int r,int rt)

    if(L<=l&&r<=R)
    
        sg[rt].val=(r-l+1)-sg[rt].val;
        sg[rt].lazy^=1;
        return;
    
    int m=(l+r)>>1;
    pushdown(rt,m-l+1,r-m);
    if(L<=m) update(L,R,l,m,rt<<1);
    if(R>m)  update(L,R,m+1,r,rt<<1|1);
    pushup(rt);

int query(int L,int R,int l,int r,int rt)

    if(L<=l&&r<=R)
        return sg[rt].val;
    if(L>r||R<l)
        return 0;
    int m=(l+r)/2;
    pushdown(rt,m-l+1,r-m);
    return query(L,R,l,m,rt<<1)+query(L,R,m+1,r,rt<<1|1);

signed main()

    scanf("%lld%lld",&n,&m);
    string s;cin>>s;
    for(int i=1;i<=n;i++)
    
        if(s[i-1]=='1')
            a[i]=1;
        else
            a[i]=0;
    
    build(1,n,1);
    while(m--)
    
       int g,x,y;scanf("%lld%lld%lld",&g,&x,&y);
       if(g==0)
       
           update(x,y,1,n,1);
       
       else if(g==1)
       
           cout<<query(x,y,1,n,1)<<endl;
       
    
    return 0;

https://www.luogu.com.cn/problem/P2880
线段树找区间极大值和极小值相减

#include <bits/stdc++.h>

using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=2e5+5;
int n,m,a[maxn];
struct node

    int val,v1=inf;
st[maxn<<2];
void pushup(int rt)

    st[rt].val=max(st[rt<<1].val,st[rt<<1|1].val);
    st[rt].v1=min(st[rt<<1].v1,st[rt<<1|1].v1);

void build(int l,int r,int rt)

    if(l==r)
    
        st[rt].val=a[l];
        st[rt].v1=a[l];
        return;
    
    int m=(l+r)>>1;
    build(l,m,rt<<1);
    build(m+1,r,rt<<1|1);
    pushup(rt);

int query(int L,int R,int l,int r,int rt)  //查询极大值

    if(L<=l&&r<=R)
        return st[rt].val;
    if(L>r||R<l)
        return 0;
    int mid=(l+r)>>1,ans=0;
    if(L<=mid)
        ans=max(ans,query(L,R,l,mid,rt<<1));
    if(R>mid)
        ans=max(ans,query(L,R,mid+1,r,rt<<1|1));
    return ans;

int query1(int L,int R,int l,int r,int rt)  //查询极小值

    if(L<=l&&r<=R)
        return st[rt].v1;
    if(L>r||R<l)
        return 0;
    int mid=(l+r)>>1,ans=inf;
    if(L<=mid)
        ans=min(ans,query1(L,R,l,mid,rt<<1));
    if(R>mid)
        ans=min(ans,query1(L,R,mid+1,r,rt<<1|1));
    return ans;

int main()

    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    build(1,n,1);
    while(m--)
    
        int x,y;scanf("%d%d",&x,&y);
        cout<<query(x,y,1,n,1)-query1(x,y,1,n,1)<<endl;
    
    return 0;