Luogt21778 New Year, luogt21778

Luogt21778 New Year, luogt21778
Description

There are n (1 \ leq n \ leq 10 ^ 5) n (1 ≤ n ≤ 105) Children, Chinese New Year, the m (1 \ leq m \ leq 10 ^ 5) m (1 ≤ m ≤ 105) gift will be given.

Three parameters l, r, k (1 \ leq l \ leq r \ leq n, 1 \ leq k \ leq 10 ^ 5) l, r, k (1 ≤ l ≤ r ≤ n, 1 ≤ k ≤ 105) indicates that a gift kk is sent to the children in the range [l, r] [l, r.

After all the gifts are given, each of the children will receive the most frequently-received gifts. If there are multiple, the one with the smallest output number; if not, the output is-1-1.

Input/Output Format

Input Format:

 

The two integers n, mn, m in the first line indicate the meaning as described above.

Next, the mm line contains three numbers: l, r, kl, r, k. The meaning is described above.

 

Output Format:

 

A total of nn rows. Each row has one number, indicating the answer.

 

Input and Output sample input sample #1: Copy

6 31 5 12 3 23 4 2

Output example #1: Copy

11211-1


There is no way of thinking. It's all about routines.
Create a line segment tree based on the sequence number of the small pot friends
Make a difference for each query.
Mark on the tree to record the location of the maximum and maximum values
After emmm, you should consider how to write the line segment tree. I feel that using the DFS sequence is not only memory-intensive, but also easy to write.

// luogu-judger-enable-o2#include<iostream>#include<vector>#include<cstdio>using namespace std;const int MAXN=1e6+10;struct node{    int l,r,ls,rs,mx,mxpos;}T[MAXN];vector<int>v[MAXN];int root,tot;void Build(int &k , int ll , int rr){    k=tot++;    T[k].mx=0; T[k].l = ll ; T[k].r = rr;    if( ll == rr  ) { T[k].mxpos = ll; return ; }    int mid=ll + rr >>1;    Build( T[k].ls , ll , mid );    Build( T[k].rs , mid+1 , rr );}void update(int k){    if( T[ T[k].ls ].mx >= T[ T[k].rs ].mx ) T[k].mx = T[ T[k].ls ].mx , T[k].mxpos = T[ T[k].ls ].mxpos;    else                                     T[k].mx = T[ T[k].rs ].mx , T[k].mxpos = T[ T[k].rs ].mxpos;}void Add(int k, int pos ){    if( T[k].l == T[k].r )    {        T[k].mx++;        return ;    }    int mid=T[k].l + T[k].r >>1;    if(pos<=mid) Add( T[k].ls , pos );    else          Add( T[k].rs , pos );    update(k);}void Delet(int k, int pos ){    if( T[k].l == T[k].r )    {        T[k].mx--;        return ;    }    int mid= T[k].l + T[k].r >>1;    if(pos<=mid) Delet( T[k].ls , pos );    else          Delet( T[k].rs , pos );    update(k);}int main(){    #ifdef WIN32    freopen("a.in","r",stdin);    #else    #endif    int N,M;    scanf("%d%d",&N,&M);    for(int i=1; i<=M ;i++ )    {        int l,r,k;        scanf("%d%d%d",&l,&r,&k);        v[l].push_back(k);        v[r+1].push_back(-k);    }    Build(root,1,N);    for(int i=1; i<=N ;i++)    {        for(int j=0; j<v[i].size() ;j++ )        {        //    printf("*%d*",v[i][j]);            if( v[i][j]>0 )                 Add(root , v[i][j] );            if( v[i][j]<0 )                 Delet(root , -v[i][j] );         }         if( T[root].mx )              printf("%d\n",T[ root ].mxpos );        else            printf("-1\n");    }        return 0;}