[Codeforces]817F. MEX Queries 离散化+线段树维护

[Codeforces]817F. MEX Queries

You are given a set of integer numbers, initially it is empty. You should perform n queries.
There are three different types of queries:
1 l r — Add all missing numbers from the interval [l, r]
2 l r — Remove all present numbers from the interval [l, r]
3 l r — Invert the interval [l, r] — add all missing and remove all present numbers from the interval [l, r]
After each query you should output MEX of the set — the smallest positive (MEX  ≥ 1) integer number which is not presented in the set.
Input
The first line contains one integer number n (1 ≤ n ≤ 105).
Next n lines contain three integer numbers t, l, r (1 ≤ t ≤ 3, 1 ≤ l ≤ r ≤ 1018) — type of the query, left and right bounds.
Output
Print MEX of the set after each query.
Examples
input
3
1 3 4
3 1 6
2 1 3
output
1
3
1
input
4
1 1 3
3 5 6
2 4 4
3 1 6
output
4
4
4
1
Note
Here are contents of the set after each query in the first example:
{3, 4} — the interval [3, 4] is added
{1, 2, 5, 6} — numbers {3, 4} from the interval [1, 6] got deleted and all the others are added
{5, 6} — numbers {1, 2} got deleted

题意

给你一个无限长的数组，初始的时候都为0，操作1是把给定区间清零，操作2是把给定区间设为1，操作3把给定区间反转。每次操作后要输出最小位置的0。

题解

看到数据范围n<=10^5，结合题意可以考虑使用线段树维护对区间的修改操作。但是l，r<=10^18，所以首先要离散化一下。在使用线段树维护的时候，节点维护该区间数相加的总和。对于操作1和操作2,我们分别赋值为1和0，对于操作3，我们把区间反转，那么新的区间和就是区间的长度减去原来的区间和。然后每次查询最小位置的0，只需要看一下左儿子所代表的区间是否小于这个区间的长度，如果是就在左儿子，否则就在右儿子查找。

题目细节

这道题有很多坑人的点，首先，在离散化的时候必须把1也加上，因为答案可能为1；线段树在下传标记时要注意顺序；记录原来信息的数组必须得开long long，空间一定要开够。

代码

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<map>using namespace std;#define ll long long#define REP(i,a,b) for(register int i=(a),_end_=(b);i<=_end_;i++)#define DREP(i,a,b) for(register int i=(a),_end_=(b);i>=_end_;i--)#define EREP(i,a) for(register int i=start[(a)];i;i=e[i].next)inline int read(){    int sum=0,p=1;char ch=getchar();    while(!(('0'<=ch && ch<='9') || ch=='-'))ch=getchar();    if(ch=='-')p=-1,ch=getchar();    while('0'<=ch && ch<='9')sum=sum*10+ch-48,ch=getchar();    return sum*p;}const int maxn=150020;map <ll,int> mp;int m,cnt;ll s[maxn*3],n;struct qu {    ll l,r;    int type;}a[maxn];struct node {    int s,lz,id;//s记录区间和，lz为懒标记，id维护区间是否反转}c[maxn*10];#define lc (o<<1)#define rc (o<<1 | 1)#define left lc,l,mid#define right rc,mid+1,rinline void make_tree(int o,int l,int r){    c[o].s=0;c[o].lz=-1;c[o].id=0;    if(l==r)return;    int mid=(l+r)>>1;    make_tree(left);    make_tree(right);}void maintain(int o,int l,int r){    c[o].s=c[lc].s+c[rc].s;}void pushdown(int o,int l,int r){    int mid=(l+r)>>1;    if(c[o].lz!=-1)//下传懒标记，同时将儿子节点的反转标记清0    {        c[lc].lz=c[rc].lz=c[o].lz;        c[lc].s=(mid-l+1)*c[o].lz;        c[rc].s=(r-mid)*c[o].lz;        c[lc].id=c[rc].id=0;        c[o].lz=-1;    }    if(c[o].id)//将儿子节点的反转标记也反转，同时维护儿子的区间和    {        c[lc].id^=1;        c[rc].id^=1;        c[lc].s=(mid-l+1)-c[lc].s;        c[rc].s=(r-mid)-c[rc].s;        c[o].id=0;    }}inline void updates(int ql,int qr,int x,int o,int l,int r){    pushdown(o,l,r);    if(ql==l && r==qr)//把区间覆盖为x    {        c[o].s=(r-l+1)*x;        c[o].lz=x;        c[o].id=0;        return;    }    int mid=(l+r)>>1;    if(ql>mid)    {        updates(ql,qr,x,right);    }    else if(qr<=mid)    {        updates(ql,qr,x,left);    }else    {        updates(ql,mid,x,left);        updates(mid+1,qr,x,right);    }    maintain(o,l,r);}inline void updatex(int ql,int qr,int o,int l,int r){    pushdown(o,l,r);    if(ql==l && r==qr)//把区间反转    {        c[o].s=(r-l+1)-c[o].s;        c[o].id^=1;        return;    }    int mid=(l+r)>>1;    if(ql>mid)    {        updatex(ql,qr,right);    }    else if(qr<=mid)    {        updatex(ql,qr,left);    }else    {        updatex(ql,mid,left);        updatex(mid+1,qr,right);    }    maintain(o,l,r);}void init(){    m=read();    REP(i,1,m)    {        cin>>a[i].type>>a[i].l>>a[i].r;        a[i].r++;        s[++cnt]=a[i].l;        s[++cnt]=a[i].r;    }    s[++cnt]=1;//答案中可能会有1，必须加上    sort(s+1,s+cnt+1);    n=unique(s+1,s+cnt+1)-(s+1);    REP(i,1,n)mp[s[i]]=i;    make_tree(1,1,n);}void query(int o,int l,int r){    if(l==r)    {        cout<<s[l]<<endl;        return;    }    int mid=(l+r)>>1;    pushdown(o,l,r);    if(c[lc].s<mid-l+1)        query(left);    else query(right);}void doing(){    REP(i,1,m)    {        int type=a[i].type,l=mp[a[i].l],r=mp[a[i].r]-1;        if(type==1)        {            updates(l,r,1,1,1,n);        }        else if(type==2)        {            updates(l,r,0,1,1,n);        }else        {            updatex(l,r,1,1,n);        }        query(1,1,n);    }}int main(){    init();    doing();    return 0;}