P2572 [SCOI2010]序列操作

没什么好说的，细节题
注释放代码里

#include
using namespace std;
const int N=1e5+7;
template 
inline void read(I &x){
    int f=1;
    char c;
    for(c=getchar();c<'0'||c>'9';c=getchar()) if(c=='-') f=-1;
    for(x=0;c>='0'&&c<='9';x=(x<<3)+(x<<1)+(c&15),c=getchar());
    x*=f;
}
int n,m;
bool a[N];
struct tree{
    int l=1<<30,r=0,sum=0,lm1=0,rm1=0,lm0=0,rm0=0,mx1=0,mx0=0,op1=-1;
    bool xr=0;
    tree(int l=0,int r=0,int sum=0,int lm1=0,int rm1=0,int lm0=0,int rm0=0,int mx1=0,int mx0=0,int op1=-1,bool xr=0):
    l(l),r(r),sum(sum),lm1(lm1),rm1(rm1),lm0(lm0),rm0(rm0),mx1(mx1),mx0(mx0),op1(op1),xr(xr){}
    #define l(x) t[x].l
    #define r(x) t[x].r
    #define mx1(x) t[x].mx1
    #define mx0(x) t[x].mx0
    #define sum(x) t[x].sum
    #define lm1(x) t[x].lm1
    #define rm1(x) t[x].rm1
    #define lm0(x) t[x].lm0
    #define rm0(x) t[x].rm0
    #define op1(x) t[x].op1
    #define xr(x) t[x].xr
}t[4*N];
inline void c1(int p,bool v){
    sum(p)=v*(r(p)-l(p)+1);
    lm1(p)=v*(r(p)-l(p)+1);
    rm1(p)=v*(r(p)-l(p)+1);
    mx1(p)=v*(r(p)-l(p)+1);
    lm0(p)=(v^1)*(r(p)-l(p)+1);
    rm0(p)=(v^1)*(r(p)-l(p)+1);
    mx0(p)=(v^1)*(r(p)-l(p)+1);
    op1(p)=v;
    xr(p)=0;
    return;
}
inline void c2(int p){
    sum(p)=r(p)-l(p)+1-sum(p);
    swap(lm1(p),lm0(p));
    swap(rm1(p),rm0(p));
    swap(mx1(p),mx0(p));
    xr(p)^=1;//注意异或
    if(op1(p)!=-1) op1(p)^=1,xr(p)=0;
    return;
}
inline void pd(int p){
    if(l(p)==r(p)) return;
    if(op1(p)!=-1){
        c1(p*2,op1(p));
        c1(p*2+1,op1(p));
        op1(p)=-1;
    }
    if(xr(p)){
        c2(p*2);
        c2(p*2+1);
        xr(p)=0;
        return;
    }
}
inline void upd(int p){
    sum(p)=sum(p*2)+sum(p*2+1);
    if(lm1(p*2)==r(p*2)-l(p*2)+1) lm1(p)=lm1(p*2)+lm1(p*2+1);
        else lm1(p)=lm1(p*2);
    if(rm1(p*2+1)==r(p*2+1)-l(p*2+1)+1) rm1(p)=rm1(p*2)+rm1(p*2+1);
        else rm1(p)=rm1(p*2+1);
    if(lm0(p*2)==r(p*2)-l(p*2)+1) lm0(p)=lm0(p*2)+lm0(p*2+1);
        else lm0(p)=lm0(p*2);
    if(rm0(p*2+1)==r(p*2+1)-l(p*2+1)+1) rm0(p)=rm0(p*2)+rm0(p*2+1);
        else rm0(p)=rm0(p*2+1);
    mx0(p)=max(mx0(p*2),max(mx0(p*2+1),rm0(p*2)+lm0(p*2+1)));
    mx1(p)=max(mx1(p*2),max(mx1(p*2+1),rm1(p*2)+lm1(p*2+1)));
    return;
}
void build(int p,int l,int r){
    l(p)=l,r(p)=r,op1(p)=-1;
    if(l==r){
        c1(p,a[l]);
        return;
    }
    int mid=(l+r)/2;
    build(p*2,l,mid);
    build(p*2+1,mid+1,r);
    upd(p);
    return;
}
void change1(int p,int l,int r,bool v){
    if(l(p)>=l&&r(p)<=r){
        c1(p,v);
        return;
    }
    pd(p);
    int mid=(l(p)+r(p))/2;
    if(mid>=l) change1(p*2,l,r,v);
    if(mid+1<=r) change1(p*2+1,l,r,v);
    upd(p);
    return;
}
void change2(int p,int l,int r){
    if(l(p)>=l&&r(p)<=r){
        c2(p);
        return;
    }
    pd(p);
    int mid=(l(p)+r(p))/2;
    if(mid>=l) change2(p*2,l,r);
    if(mid=l&&r(p)<=r){
        return sum(p);
    }
    pd(p);
    int res=0;
    int mid=(l(p)+r(p))/2;
    if(mid>=l) res+=ask1(p*2,l,r);
    if(mid=l&&r(p)<=r){
        return t[p];
    }
    pd(p);
    tree res,res1,res2;
    int mid=(l(p)+r(p))/2;
    if(mid>=l) res1=ask2(p*2,l,r);
    if(mid=r) return res1;
    res.l=res1.l,res.r=res2.r;
    if(res1.lm1==res1.r-res1.l+1) res.lm1=res1.lm1+res2.lm1;
        else res.lm1=res1.lm1;
    if(res2.rm1==res2.r-res2.l+1) res.rm1=res2.rm1+res1.rm1;
        else res.rm1=res2.rm1;
    res.mx1=max(res1.mx1,max(res2.mx1,res1.rm1+res2.lm1));
    return res;
}
int main(){
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    read(n),read(m);
    for(int i=1;i<=n;i++)
        read(a[i]);
    build(1,1,n);
    for(int i=1;i<=m;i++){
        int op,A,b;
        read(op),read(A),read(b);
        A++,b++;
        if(op==0) change1(1,A,b,0);
        if(op==1) change1(1,A,b,1);
        if(op==2) change2(1,A,b);
        if(op==3) printf("%d\n",ask1(1,A,b));
        if(op==4) printf("%d\n",ask2(1,A,b).mx1);
    }
    return 0;
}