#include <bits/stdc++.h>
using namespace std;

const int N = 4009;
int n, m;
double k;
struct task
{
    int value, power, heat;
    int res, id;
} ta[N];
struct fuwu
{
    int p;
    double h;
    double curh;
    int curp;
} fu[N];

bool cmp1(task t1, task t2)
{
    if (t1.value == t2.value)
    {
        if (t1.power == t2.power)
            return t1.heat <= t2.heat;
        else
            return t1.power < t2.power;
    }
    else
        return t1.value > t2.value;
}
bool cmp2(task t1, task t2)
{
    return t1.id < t2.id;
}

bool check(int i, int j)
{
    double t1, t2, t3;
    t2 = fu[j].curh + ta[i].heat;
    if (t2 > fu[j].h)
        return 0;
    if (j == 1)
    {
        t3 = fu[j + 1].curh + ta[i].heat * k;
        if (t3 > fu[j + 1].h)
            return 0;
    }
    else if (j == m)
    {
        t1 = fu[j - 1].curh + ta[i].heat * k;
        if (t1 > fu[j - 1].h)
            return 0;
    }
    else
    {
        t3 = fu[j + 1].curh + ta[i].heat * k;
        t1 = fu[j - 1].curh + ta[i].heat * k;
        if (t1 > fu[j - 1].h)
            return 0;
        if (t3 > fu[j + 1].h)
            return 0;
    }
    return 1;
}

void renew(int j, int i)
{
    fu[j].curp += ta[i].power;
    fu[j].curh += ta[i].heat;
    if (j == 1)
        fu[j + 1].curh += ta[i].heat * k;
    else if (j == m)
        fu[j - 1].curh += ta[i].heat * k;
    else
    {
        fu[j + 1].curh += ta[i].heat * k;
        fu[j - 1].curh += ta[i].heat * k;
    }
}

int main()
{
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++)
    {
        cin >> ta[i].value >> ta[i].power >> ta[i].heat;
        ta[i].id = i;
    }
    for (int i = 1; i <= m; i++)
    {
        cin >> fu[i].p >> fu[i].h;
        fu[i].curh = fu[i].curp = 0;
    }

    sort(ta + 1, ta + n + 1, cmp1);
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            if (fu[j].curp + ta[i].power <= fu[j].p && check(i, j))
            {
                renew(j, i);
                ta[i].res = j;
                // cout << j << endl;
                break;
            }
            else
                ta[i].res = 0;
        }
        // cout << ta[i].id << " " << ta[i].res << endl;
    }
    sort(ta + 1, ta + n + 1, cmp2);
    /// for (int i = 1; i <= m; i++) cout << fu[i].curh << endl;
    for (int i = 1; i <= n; i++)
    {
        cout << ta[i].res << " ";
    }
}

考试题目大意时在满足每个机器的发热不超过阈值的情况下保证最终结果最优，此代码得分大概370多分，能够保证学生在算法巅峰赛得到国一水平。