CodeForce 356A Knight Tournament（线段树的区间更新+单点询问）

ゞ浴缸里的玫瑰 2022-06-10 05:17 164阅读 0赞

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 *l**i* and at most *r**i* have 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 *x**i*, he continued participating in the tournament. Other knights left the tournament.
 *  The winner of the last (the *m*\-th) fight (the knight number *x**m*) 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 *l**i*, *r**i*, *x**i* (1 ≤ *l**i* < *r**i* ≤ *n*; *l**i* ≤ *x**i* ≤ *r**i*) — 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.

Example

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.

题解：

题意：

有n个骑士，m场战斗，每场战斗有l，r，x，表示下标为l到r的骑士中胜者为x，最后让你输出每个人被谁打败了，如果是人生大赢家就输出0

思路：

比赛一开始以为是一个不路径压缩的并查集，后来一想这么大的区间更新肯定要超时就想着线段树了，把要更新的区间倒过来进行区间覆盖就好了（注意去掉x）

代码：

#include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<stdio.h>
    #include<math.h>
    #include<string>
    #include<stdio.h>
    #include<queue>
    #include<stack>
    #include<map>
    #include<deque>
    #define M (t[k].l+t[k].r)/2
    #define lson k*2
    #define rson k*2+1
    #define ll long long
    #define INF 100861111;
    using namespace std;
    struct node
    {
        int l,r;
        int tag;//lazy tag
        int v;//值
    }t[300005*4];
    void Build(int l,int r,int k)
    {
        t[k].l=l;
        t[k].r=r;
        t[k].tag=0;
        t[k].v=0;
        if(l==r)
            return;
        int mid=M;
        Build(l,mid,lson);
        Build(mid+1,r,rson);
    }
    void pushdown(int k)//更新子区间，下推标记
    {
        if(t[k].tag)
        {
            t[lson].tag=t[rson].tag=t[k].tag;
            t[lson].v=t[rson].v=t[k].tag;
            t[k].tag=0;
        }
    }
    void update(int l,int r,int k,int v)//区间更新
    {
        if(t[k].l==l&&t[k].r==r)
        {
            t[k].v=v;
            t[k].tag=v;
            return;
        }
        pushdown(k);
        int mid=M;
        if(r<=mid)
        {
            update(l,r,lson,v);
        }
        else if(l>mid)
            update(l,r,rson,v);
        else
        {
            update(l,mid,lson,v);
            update(mid+1,r,rson,v);
        }
    }
    int query(int x,int k)//单点询问
    {
        if(t[k].l==t[k].r)
        {
            return t[k].v;
        }
        pushdown(k);
        int mid=M;
        if(x<=mid)
            return query(x,lson);
        else
            return query(x,rson);
    }
    struct edge
    {
        int l,r;
        int x;
    }q[300005];//存更新区间和信息
    int main()
    {
        int i,j,k,n,m;
        scanf("%d%d",&n,&m);
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].x);
            if(q[i].l>q[i].r)//可能有坑
                swap(q[i].l,q[i].r);
        }
        Build(1,n,1);
        for(i=m-1;i>=0;i--)//反过来处理更新
        {
            if(q[i].l!=q[i].x)//如果不是胜者在最右边的情况
            update(q[i].l,q[i].x-1,1,q[i].x);
            if(q[i].r!=q[i].x)//如果不是胜者在最左边的情况
            update(q[i].x+1,q[i].r,1,q[i].x);
        }
        int temp=query(1,1);//特殊处理第一个人
        if(temp==1)
            printf("0");
        else
            printf("%d",temp);
        for(i=2;i<=n;i++)//处理后面的人
        {
            int temp=query(i,1);
            if(temp==i)
                printf(" 0");
            else
                printf(" %d",temp);
        }
        printf("\n");
        return 0;
    }