利用 stl 来做就可以了，在此发现 algortihm 的 lower_bound 不及 set 的
貌似 set 里面已经有做处理了
/*
* GCA : "Computer is artificial subject absolutely,Math is God"
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define VAR(a,b) __typeof(b) a=(b)
#define debug(...) printf("DEBUG: "),printf(__VA_ARGS__)
#else
#define VAR(a,b) __typeof(b) a=(b)
#define debug(...)
#endif
typedef unsigned int uint;
typedef long long int Int;
typedef unsigned long long int UInt;
#define Set(a,s) memset(a,s,sizeof(a))
#define Pln() printf("\n")
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define M 300005
#define PB push_back
#define oo INT_MAX
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)
#define eps 1e-9
#define X first
#define Y second
inline bool xdy(double x,double y){return x>y+eps;}
inline bool xddy(double x,double y){return x>y-eps;}
inline bool xcy(double x,double y){return x<y-eps;}
inline bool xcdy(double x,double y){return x<y+eps;}
const Int mod=1000000007;
int ans[M];
int n,m;
set<int>::iterator a,b,it;
set<int> s;
int main() {
ios_base::sync_with_stdio(0);
while(~scanf("%d%d",&n,&m)){
for(int i=1;i<=n;i++)s.insert(i);
while(m--){
int l,r,k;
scanf("%d%d%d",&l,&r,&k);
a=s.lower_bound(l);
b=s.upper_bound(r);
for(it=a;it!=b;it++){
if(!ans[*it]&&*it!=k)ans[*it]=k;
}
s.erase(a,b);
s.insert(k);
}
for(int i=1;i<=n;i++){
printf("%d ",ans[i]);
}
}
}