Codeforces 482B. Interesting Array 线段树

B. Interesting Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value  should be equal to qi.

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 li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 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 test(s)

input

3 1
1 3 3

output

YES
3 3 3

input

3 2
1 3 3
1 3 2

output

NO

#include 
#include 
#include 
#include 
#include 

using namespace std;

const int maxn=100005;

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

int a[maxn<<2],n,m;
int OR[maxn<<2];

struct IA
{
    int l,r,q;
}ia[maxn];

void push_up(int rt)
{
    a[rt]=a[rt<<1]&a[rt<<1|1];
}

void push_down(int rt)
{
    if(OR[rt])
    {
        a[rt<<1]|=OR[rt]; a[rt<<1|1]|=OR[rt];
        OR[rt<<1]|=OR[rt]; OR[rt<<1|1]|=OR[rt];
        OR[rt]=0;
    }
}

void update(int L,int R,int C,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        OR[rt]|=C;
        a[rt]|=C;
        return ;
    }
    push_down(rt);
    int m=(l+r)/2;
    if(L<=m) update(L,R,C,lson);
    if(R>m) update(L,R,C,rson);
    push_up(rt);
}

int query(int L,int R,int l,int r,int rt)
{
   if(L<=l&&r<=R)
   {
       return a[rt];
   }
   int m=(l+r)/2;
   push_down(rt);
   int part1=-1,part2=-1;
   if(L<=m) part1=query(L,R,lson);
   if(R>m) part2=query(L,R,rson);
   if(part1==-1) return part2;
   if(part2==-1) return part1;
   return part1&part2;
}

vector ans;

void showit(int l,int r,int rt)
{
    if(l==r)
    {
        ans.push_back(a[rt]);
        return ;
    }
    int m=(l+r)/2;
    push_down(rt);
    showit(lson);
    showit(rson);
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i

转载于:https://www.cnblogs.com/mfmdaoyou/p/6752082.html