【codevs2421】【BZOJ1858】序列操作，线段树

传送门1

传送门2

写在前面：垃圾题目毁我青春

思路：

这是一道很明显但你就是要调很久而且不会舒服的线段树

每个区间你需要维护7个量和2个标记

分别是1的总数，最长连续1(0)长度，左(右)端点的连续1(0)的长度，翻转标记和01标记

数量的维护比较容易，可以参考seating，关键是标记

翻转标记是会影响当前的01标记，比如当前区间全是1（标记为1），那么翻转后01标记要变为0。所以下放标记时要时刻小心，最好先翻转再01，而且如果当前区间内全为0或1，下放时就可以把左右儿子的翻转标记清零了

翻转操作时，1的总数变为区间大小-原来1的总数，剩下的6个量直接两两交换

查询连续1的长度时，注意左右儿子的区间合并时，需要获得左儿子右端1和右儿子左端1的合并长度，这里不能直接加，而是还要和查询区间[L,R]的长度比较，取他们的较小值

注意：一定要打两个标记！我当初作死只打一个，调了好久！

#include<bits/stdc++.h>
#define M 100007
using namespace std;
int n,m,a[M];
int in()
{
int t=0;
char ch=getchar();
while (ch<'0'||ch>'9') ch=getchar();
while (ch>='0'&&ch<='9') t=(t<<3)+(t<<1)+ch-'0',ch=getchar();
return t;
}
struct os
{
int sum,L_1,R_1,L_0,R_0,mx_0,mx_1,lazy;
bool rev;
void swaps()
{
swap(L_1,L_0);
swap(R_1,R_0);
swap(mx_0,mx_1);
}
void put(int a,int b,int c,int d,int e,int f,int g,int h,bool i)
{
sum=a;L_1=b;R_1=c;
L_0=d;R_0=e;mx_0=f;
mx_1=g;lazy=h;rev=i;
}
}tree[M<<2];
void pushup(int x,int begin,int end)
{
int mid=(begin+end)>>1;
tree[x].sum=tree[x<<1].sum+tree[x<<1|1].sum;
tree[x].L_1=tree[x<<1].L_1;
if (tree[x<<1].L_1==mid-begin+1)tree[x].L_1+=tree[x<<1|1].L_1;
tree[x].L_0=tree[x<<1].L_0;
if (tree[x<<1].L_0==mid-begin+1)tree[x].L_0+=tree[x<<1|1].L_0;
tree[x].R_1=tree[x<<1|1].R_1;
if (tree[x<<1|1].R_1==end-mid)tree[x].R_1+=tree[x<<1].R_1;
tree[x].R_0=tree[x<<1|1].R_0;
if (tree[x<<1|1].R_0==end-mid)tree[x].R_0+=tree[x<<1].R_0;
tree[x].mx_0=max(tree[x<<1].R_0+tree[x<<1|1].L_0,max(tree[x<<1].mx_0,tree[x<<1|1].mx_0));
tree[x].mx_1=max(tree[x<<1].R_1+tree[x<<1|1].L_1,max(tree[x<<1].mx_1,tree[x<<1|1].mx_1));
}
void pushdown(int x,int begin,int end)
{
if (tree[x].lazy==-1&&!tree[x].rev) return;
int mid=(begin+end)>>1;
if (tree[x].rev)
{
tree[x<<1].rev^=1;
if (tree[x<<1].lazy!=-1)tree[x<<1].lazy^=1;
tree[x<<1].sum=mid-begin+1-tree[x<<1].sum;
tree[x<<1].swaps();
tree[x<<1|1].rev^=1;
if (tree[x<<1|1].lazy!=-1) tree[x<<1|1].lazy^=1;
tree[x<<1|1].sum=end-mid-tree[x<<1|1].sum;
tree[x<<1|1].swaps();
}
if (tree[x].lazy==1)
{
tree[x<<1].put(mid-begin+1,mid-begin+1,mid-begin+1,0,0,0,mid-begin+1,1,0);
tree[x<<1|1].put(end-mid,end-mid,end-mid,0,0,0,end-mid,1,0);
}
else if (tree[x].lazy==0)
{
tree[x<<1].put(0,0,0,mid-begin+1,mid-begin+1,mid-begin+1,0,0,0);
tree[x<<1|1].put(0,0,0,end-mid,end-mid,end-mid,0,0,0);
}
tree[x].lazy=-1;tree[x].rev=0;
}
void build(int x,int begin,int end)
{
if (begin==end)
{
if (a[end]) tree[x].put(1,1,1,0,0,0,1,-1,0);
else tree[x].put(0,0,0,1,1,1,0,-1,0);
return;
}
tree[x].lazy=-1;
int mid=(begin+end)>>1;
build(x<<1,begin,mid);
build(x<<1|1,mid+1,end);
pushup(x,begin,end);
}
void update(int x,int begin,int end,int l,int r,int num)
{
if (l<=begin&&end<=r)
{
if (num)
tree[x].put(end-begin+1,end-begin+1,end-begin+1,0,0,0,end-begin+1,1,0);
else tree[x].put(0,0,0,end-begin+1,end-begin+1,end-begin+1,0,0,0);
return;
}
pushdown(x,begin,end);
int mid=(begin+end)>>1;
if (l<=mid) update(x<<1,begin,mid,l,r,num);
if (r>mid) update(x<<1|1,mid+1,end,l,r,num);
pushup(x,begin,end);
}
void oppo(int x,int begin,int end,int l,int r)
{
if (l<=begin&&end<=r)
{
tree[x].swaps();
tree[x].sum=end-begin+1-tree[x].sum;
if (tree[x].lazy!=-1) tree[x].lazy^=1;
tree[x].rev^=1;
return;
}
pushdown(x,begin,end);
int mid=(begin+end)>>1;
if (l<=mid) oppo(x<<1,begin,mid,l,r);
if (r>mid) oppo(x<<1|1,mid+1,end,l,r);
pushup(x,begin,end);
}
int get_sum(int x,int begin,int end,int l,int r)
{
if (l<=begin&&end<=r) return tree[x].sum;
pushdown(x,begin,end);
int mid=(begin+end)>>1,ans=0;
if (l<=mid) ans+=get_sum(x<<1,begin,mid,l,r);
if (r>mid) ans+=get_sum(x<<1|1,mid+1,end,l,r);
return ans;
}
int get_con(int x,int begin,int end,int l,int r)
{
if (l<=begin&&end<=r) return tree[x].mx_1;
pushdown(x,begin,end);
int mid=(begin+end)>>1,ans=0;
if (mid>=l) ans=max(ans,get_con(x<<1,begin,mid,l,r));
if (mid<r) ans=max(ans,get_con(x<<1|1,mid+1,end,l,r));
if (l<=mid&&mid<r)
ans=max(ans,min(tree[x<<1].R_1,mid-l+1)+min(tree[x<<1|1].L_1,r-mid));
return ans;
}
main()
{
n=in();m=in();
for (int i=1;i<=n;i++) a[i]=in();
build(1,1,n);
int opt,x,y;
while (m--)
{
opt=in();x=in()+1;y=in()+1;
if (opt==0||opt==1) update(1,1,n,x,y,opt);
else if (opt==2) oppo(1,1,n,x,y);
else if (opt==3) printf("%d\n",get_sum(1,1,n,x,y));
else printf("%d\n",get_con(1,1,n,x,y));
}
}