线段树之区间更新Interesting Array

Description

We'll call an array of n non-negative integers a[1], a[2], ..., a[n]interesting, if it meets m constraints. The i-th of the m constraints consists of three integers l_i, r_i, q_i (1 ≤ l_i ≤ r_i ≤ n) meaning that value  should be equal to q_i.

Your task is to find any interesting array of n elements or state that such array doesn't exist.

Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".

Input

The first line contains two integers n, m (1 ≤ n ≤ 10⁵, 1 ≤ m ≤ 10⁵) — the number of elements in the array and the number of limits.

Each of the next m lines contains three integers l_i, r_i, q_i (1 ≤ l_i ≤ r_i ≤ n, 0 ≤ q_i < 2³⁰) describing the i-th limit.

Output

If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 2³⁰) decribing the interesting array. If there are multiple answers, print any of them.

If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.

Sample Input

Input

3 1
1 3 3

Output

YES
3 3 3

Input

3 2
1 3 3
1 3 2

Output

NO

#include<cstdio>
using namespace std;

const int N=1e5+5;
int f[N<<2],lazy[N<<2],le[N],ri[N],q[N];

void build(int root,int left,int right){
    
    if(left==right){
        f[root]=lazy[root]=0;
        return;
    }
    
    int mid,rt;
    
    mid=(left+right)>>1;
    rt=root<<1;
    
    build(rt,left,mid);
    build(rt+1,mid+1,right);
}

void pushdown(int root){
    int rt=root<<1;
    
    lazy[rt]|=lazy[root];
    lazy[rt+1]|=lazy[root];
    
    f[rt]|=lazy[root];
    f[rt+1]|=lazy[root];
    
    lazy[root]=0;
}
void update(int root,int left,int right,int uleft,int uright,int val){
    if(uleft<=left&&right<=uright){
        f[root]|=val;
        lazy[root]|=val;
        return;
    }
    
    if(lazy[root]!=0){
        pushdown(root);
    }
    
    int mid,rt;
    mid=(left+right)>>1;
    rt=root<<1;
    
    if(uleft<=mid){
        update(rt,left,mid,uleft,uright,val);
    }
    if(uright>mid){
        update(rt+1,mid+1,right,uleft,uright,val);
    }
    
    f[root]=f[rt]&f[rt+1];
}

int query(int root,int left,int right,int qleft,int qright){
    if(qleft<=left&&right<=qright){
        return f[root];
    }
    
    if(lazy[root]!=0){
        pushdown(root);
    }
    
    int rt,mid,ans;
    ans=-1;
    rt=root<<1;
    mid=(left+right)>>1;
    
    if(qleft<=mid){
        ans&=query(rt,left,mid,qleft,qright);
    }
    if(qright>mid){
        ans&=query(rt+1,mid+1,right,qleft,qright);
    }
    return ans;

}
int main(){
    int n,m,flag;
    
    scanf("%d%d",&n,&m);
    
    for(int i=1;i<=m;i++){
        
        scanf("%d%d%d",&le[i],&ri[i],&q[i]);
        
        update(1,1,n,le[i],ri[i],q[i]);
    }
    
    flag=1;
    
    for(int i=1;i<=m;i++){
        if(query(1,1,n,le[i],ri[i])!=q[i]){
            flag=0;
            break;
        }
    }
    
    if(flag==1){
        printf("YES\n");
        for(int i=1;i<=n;i++){
            printf("%d ",query(1,1,n,i,i));
        }
    }else{
        printf("NO\n");
    }
    return 0;
}

 

标签：rt,integers,int,Interesting,线段,interesting,qi,Array,array
来源： https://www.cnblogs.com/killjoyskr/p/16492275.html