Gym - 101102C Bored Judge

原创于 2017-08-19 09:52:03 发布 · 434 阅读

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本文介绍了一个算法问题，旨在通过分析比赛日志文件来确定冠军队伍是否稳定不变。使用线段树结构来高效更新和查询各队伍分数，以便快速判断冠军变动情况。

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Judge Bahosain was bored at ACM AmrahCPC 2016 as the winner of the contest had the first rank from the second hour until the end of the contest.

Bahosain is studying the results of the past contests to improve the problem sets he writes and make sure this won’t happen again.

Bahosain will provide you with the log file of each contest, your task is to find the first moment after which the winner of the contest doesn’t change.

The winner of the contest is the team with the highest points. If there’s more than one team with the same points, then the winner is the team with smallest team ID number.

Input

The first line of input contains a single integer T, the number of test cases.

The first line of each test case contains two space-separated integers N and Q (1 ≤ N, Q ≤ 10⁵), the number of teams and the number of events in the log file. Teams are numbered from 1 to N.

Each of the following Q lines represents an event in the form: X P, which means team number X (1 ≤ X ≤ N) got P ( - 100 ≤ P ≤ 100, P ≠ 0) points. Note that P can be negative, in this case it represents an unsuccessful hacking attempt.

Log events are given in the chronological order.

Initially, the score of each team is zero.

Output

For each test case, if the winner of the contest never changes during the contest, print 0. Otherwise, print the number of the first event after which the winner of the contest didn’t change. Log events are numbered from 1 to Q in the given order.

Example

Input

Output

这个题的题意是从哪句话开始一直到结束冠军在没有发生过变化。

这个题用线段树每次修改值的时候记录当前冠军和之前冠军进行比较如果两个冠军不一样记录当前位置。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#define inf 0x3f3f3f3f

using namespace std;

struct node
{
	int w;
	int num;
};

node p[110000<<2];
int k;

node max(node a, node b)
{
	if(a.w == b.w)
		return a.num < b.num?a:b;
	return a.w > b.w?a:b;
}

void build(int l, int r, int x)
{
	if(l == r)
	{
		p[x].w = 0;
		p[x].num = k;
		k++;
		return;
	}
	int m = (l + r) >> 1;
	build(l,m,x<<1);
	build(m+1,r,x<<1|1);
	p[x] = max(p[x<<1],p[x<<1|1]);
}

void Updata(int i, int j, int l, int r, int x)
{
	if(l == r)
	{
		p[x].w += j;
		return;
	}
	int m = (l + r) >> 1;
	if(i <= m)
		Updata(i,j,l,m,x<<1);
	else
		Updata(i,j,m+1,r,x<<1|1);
	p[x] = max(p[x<<1],p[x<<1|1]);
}

node Query(int i, int j, int l, int r, int x)
{
	if(l >= i && r <= j)
	{
		return p[x];
	}
	node a;
	a.w = 0;
	a.num = inf;
	int m = (l + r) >> 1;
	if(i <= m)
		a = max(a,Query(i,j,l,m,x<<1));
	if(i > m)
		a = max(a,Query(i,j,m+1,r,x<<1|1));
    return a;
}

int main()
{
	int t, n, m, i, j, x, y, l;
	node a;
	scanf("%d",&t);
	while(t--)
	{
		k = 1;
		scanf("%d %d",&n, &m);
		x = 0;
		y = 1;
		build(1,n,1);
		for(l = 1;l <= m;l++)
		{
			scanf("%d %d",&i, &j);
			Updata(i,j,1,n,1);
			a = Query(1,n,1,n,1);
			if(y != a.num)
			{
				x = l;
				y = a.num;
			}
		}
		printf("%d\n",x);
	}
	return 0;
}