CodeForces 356A_(set应用，线段树)

A. Knight Tournament

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Hooray! Berl II, the king of Berland is making a knight tournament. The king has already sent the message to all knights in the kingdom and they in turn agreed to participate in this grand event.

As for you, you're just a simple peasant. There's no surprise that you slept in this morning and were late for the tournament (it was a weekend, after all). Now you are really curious about the results of the tournament. This time the tournament in Berland went as follows:

• There are n knights participating in the tournament. Each knight was assigned his unique number — an integer from 1 to n.
• The tournament consisted of m fights, in the i-th fight the knights that were still in the game with numbers at least li and at most rihave fought for the right to continue taking part in the tournament.
• After the i-th fight among all participants of the fight only one knight won — the knight number xi, he continued participating in the tournament. Other knights left the tournament.
• The winner of the last (the m-th) fight (the knight number xm) became the winner of the tournament.

You fished out all the information about the fights from your friends. Now for each knight you want to know the name of the knight he was conquered by. We think that the knight number b was conquered by the knight number a, if there was a fight with both of these knights present and the winner was the knight number a.

Write the code that calculates for each knight, the name of the knight that beat him.

Input

The first line contains two integers n, m (2 ≤ n ≤ 3·105; 1 ≤ m ≤ 3·105) — the number of knights and the number of fights. Each of the following m lines contains three integers li, ri, xi (1 ≤ li < ri ≤ n; li ≤ xi ≤ ri) — the description of the i-th fight.

It is guaranteed that the input is correct and matches the problem statement. It is guaranteed that at least two knights took part in each battle.

Output

Print n integers. If the i-th knight lost, then the i-th number should equal the number of the knight that beat the knight number i. If the i-th knight is the winner, then the i-th number must equal 0.

Examples

input

4 3
1 2 1
1 3 3
1 4 4

output

3 1 4 0 

input

8 4
3 5 4
3 7 6
2 8 8
1 8 1

output

0 8 4 6 4 8 6 1 

n个骑士，m场战斗，知道m场战斗的（l,r,x）（从l到r中未出局的骑士中x胜出），求每个骑士被谁战胜。

1）.STL_SET 的应用

#include <stdio.h>
#include <string.h>
#include <math.h>
#include<iostream>
#include <algorithm>
#include<set>
using namespace std;
#define MAXN 300005
set<int>s;
set<int>::iterator it_l,it,tmp[MAXN];

int ans[MAXN];
int main()
{
    int n,m,l,r,x,num;
    while(cin>>n>>m)
    {
        s.clear();
        memset(ans,,sizeof(ans));
        for(int i=;i<=n;i++)
            s.insert(i);
        for(int i=;i<m;i++)
        {
            cin>>l>>r>>x;
            it_l=s.lower_bound(l);
            num=;
            for(it=it_l;*it<=r&&it!=s.end();it++)
            {
                if(*it!=x)
                {
                    ans[*it]=x;
                    tmp[num++]=it;
                }
            }
            for(int j=;j<num;j++)
                s.erase(tmp[j]);
        }
        for(int i=;i<n;i++)
            cout<<ans[i]<<" ";
        cout<<ans[n]<<endl;
    }
    return ;
}

2）.线段树。最开始用单点更新超时，看题解说是倒着区间更新。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include<iostream>
#include <algorithm>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1

int n,m;
struct Node
{
    int lazy;
    int l,r;
    int con;
} tree[<<];

void pushdown(int rt)
{
    if(tree[rt].lazy>)
    {
        tree[rt<<].lazy=tree[rt<<|].lazy=tree[rt].lazy;
        tree[rt<<].con=tree[rt<<|].con=tree[rt].lazy;
        tree[rt].lazy=;
    }
}

void build(int l,int r,int rt)
{
    tree[rt].l=l;
    tree[rt].r=r;
    tree[rt].lazy=;
    tree[rt].con=;
    if(l==r)
        return;
    int mid=(l+r)/;
    build(lson);
    build(rson);
}

int query(int x,int l,int r,int rt)
{
    if(l==r)
    {
        return tree[rt].lazy;
    }
    pushdown(rt);
    int mid=(r+l)>>;
    if(x>mid)
        query(x,rson);
    else
        query(x,lson);
}

void update(int L,int R,int l,int r,int rt,int con)
{
    if(L==l&&R==r)
    {
        tree[rt].lazy=con;
        return;
    }
    pushdown(rt);
    int mid=(l+r)/;
    if(L>mid)
        update(L,R,rson,con);
    else if(R<=mid)
        update(L,R,lson,con);
    else
    {
        update(L,mid,lson,con);
        update(mid+,R,rson,con);
    }
}

int l[],r[],con[];

int main()
{

    scanf("%d%d",&n,&m);
    build(,n,);
    for(int i=; i<m; i++)
        scanf("%d%d%d",&l[i],&r[i],&con[i]);
    for(int i=m-;i>=;i--)
        {
            if(con[i]>l[i])
                update(l[i],con[i]-,,n,,con[i]);
            if(con[i]<r[i])
                update(con[i]+,r[i],,n,,con[i]);
        }
    for(int i=; i<=n; i++)
    {
        printf("%d",query(i,,n,));
        if(i==n)
            printf("\n");
        else
            printf(" ");
    }
    return ;
}