Codeforces Round #207 (Div. 2) C. Knight Tournament

C. 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.

Sample test(s)

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 

Note

Consider the first test case. Knights 1 and 2 fought the first fight and knight 1 won. Knights 1 and 3 fought the second fight and knight 3 won. The last fight was between knights 3 and 4, knight 4 won.

题目大意：一堆人PK，给你一个区间和胜利者，求解每个人是被谁打败的。

题目分析：首先想到是模拟，算时间复杂度O(n2)，绝对要TLE，然后很自然的想到是线段树，但是又和一般的线段树不同，首先建树很简单，不再赘述。主要是更新，首先我们想到正着来更新，但是正着更新有个很复杂的地方就是之前出现的值是不能被覆盖的，这就导致要各种标记。接着我们想到反着就可以覆盖了，而且是必须覆盖。但是覆盖还有一个很重要的地方就是赢的那个人的信息是由前一组决定的，所以更新要分两段update(l,x-1,x)和update(x+1,r,x)。既然要用这种分段，就又会得到一个问题，就是：l>r的情况，特判。

AC代码：

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=1000006;
struct TR
{
    int l,r;
    int win;
    int mid(){return (l+r)>>1;}
}tr[maxn*4];
int rule;
void down(int R,int c)
{
    tr[R].win=c;
}
void pushdown(int R)
{
    if(tr[R].win)
    {
        down(R<<1,tr[R].win);
        down(R<<1|1,tr[R].win);
        tr[R].win=0;
    }
}
void build(int l,int r,int R)
{
    tr[R].l=l,tr[R].r=r;
    tr[R].win=0;
    if(l==r)
        return ;
    int mid=tr[R].mid();
    build(l,mid,R<<1);
    build(mid+1,r,R<<1|1);
}
void update(int l,int r,int c,int R)
{
    if(l>r)
        return ;///为了排除端点处的人胜利
    if(l<=tr[R].l && tr[R].r<=r)
    {
        tr[R].win=c;///down(R,c);
        return ;
    }
    pushdown(R);
    int mid=tr[R].mid();
    if(l<=mid)
        update(l,r,c,R<<1);
    if(r>=mid+1)
        update(l,r,c,R<<1|1);
}
int query(int x,int R)
{
    if(tr[R].l==tr[R].r)
        return tr[R].win;
    pushdown(R);
    int mid=tr[R].mid();
    if(x<=mid)
        return query(x,R<<1);
    return query(x,R<<1|1);
}
int da[maxn][3];
int main()
{
    int n,m;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        build(1,n,1);
        for(int i=0;i<m;i++)
            scanf("%d %d %d",&da[i][0],&da[i][1],&da[i][2]);
        rule=da[m-1][2];
        for(int i=m-1;i>=0;i--)
        {
            /// update(da[i][0],da[i][1],da[i][0],1);///虽然倒置有可以被覆盖的特点，但是赢了的人是不应该被覆盖的。
            update(da[i][0],da[i][2]-1,da[i][2],1);
            update(da[i][2]+1,da[i][1],da[i][2],1);
        }
        for(int i=1;i<=n;i++)
        {
            printf("%d ",query(i,1));
        }
        puts("");
    }
    return 0;
}