CodeForces

题目链接http://codeforces.com/problemset/problem/357/C点击打开链接

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 ri 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 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 31 2 11 3 31 4 4

output

3 1 4 0 

input

8 43 5 43 7 62 8 81 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.

可以用set乱搞。。现将所有人放入容器中 然后谁被打败就把他erase 用一个数组记录每个打败他的人

或者用并查集 注意用一个next数组储存他的下一个集合 不能直接用pre作为打败他的人 因为在过程中可能pre会因缩短路径变化 因此要多开一个

并查集代码

#include <iostream>#include <queue>#include <stdio.h>#include <stdlib.h>#include <stack>#include <limits>#include <string>#include <string.h>#include <vector>#include <set>#include <map>#include <algorithm>#include <math.h>using namespace std;int pre[333333];int nnext[333333];int ppre[333333];int findx(int x){    int r=x;    while(ppre[r]!=r)    {        r=ppre[r];    }    int i=x;int j;    while(ppre[i]!=r)    {        j=ppre[i];        ppre[i]=r;        i=j;    }    return r;}void join (int x,int y,int win){    int nnnext=nnext[y];    int temp=x;    while(temp<nnnext)    {        int mid=nnext[temp];        nnext[temp]=nnnext;        int midmid=findx(temp);        if(pre[midmid]==midmid)            pre[midmid]=win;        if(ppre[temp]!=win)            ppre[midmid]=win;        temp=mid;    }    int p1=findx(x);    int p2=findx(y);    if(p1!=p2)    {        ppre[p2]=p1;    }}int main(){    int n;int m;    scanf("%d%d",&n,&m);    for(int i=0;i<=n;i++)        ppre[i]=i,pre[i]=i,nnext[i]=i+1;    for(int i=0;i<m;i++)    {        int l,r,win;        scanf("%d%d%d",&l,&r,&win);        //if(l>r)        //swap(l,r);        join(l,r,win);    }    for(int i=1;i<=n;i++)        if(pre[i]!=i)            printf("%d ",pre[i]);        else            printf("0 ");}

乱搞代码

#include <iostream>#include <queue>#include <stdio.h>#include <stdlib.h>#include <stack>#include <limits>#include <string>#include <string.h>#include <vector>#include <set>#include <map>#include <algorithm>#include <math.h>using namespace std;set<int> s;set <int > ::iterator it,re[333333];int a[333333];int main(){    int n,m;    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)        s.insert(i);    for(int i=1;i<=m;i++)    {        int l,r,win;        int step=0;        scanf("%d%d%d",&l,&r,&win);        for(it=s.lower_bound(l);(*it)<=r&&it!=s.end();it++)        {            if((*it)!=win)            {                re[step++]=it;                a[*it]=win;            }        }        for(int ii=0;ii<step;ii++)            s.erase(re[ii]);    }    for(int i=1;i<=n;i++)        {            printf("%d ",a[i]);        }}