Codeforces 609D 贪心+二分

D. Gadgets for dollars and pounds

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Nura wants to buy k gadgets. She has onlys burles for that. She can buy each gadget for dollars or for pounds. So each gadget is selling only for some type of currency. The type of currency and the cost in that currency are not changing.

Nura can buy gadgets for n days. For each day you know the exchange rates of dollar and pound, so you know the cost of conversion burles to dollars or to pounds.

Each day (from 1 to n) Nura can buy some gadgets by current exchange rate. Each day she can buy any gadgets she wants, but each gadget can be bought no more than once duringn days.

Help Nura to find the minimum day index when she will have k gadgets. Nura always pays with burles, which are converted according to the exchange rate of the purchase day. Nura can't buy dollars or pounds, she always stores only burles. Gadgets are numbered with integers from1 to m in order of their appearing in input.

Input

First line contains four integers n, m, k, s (1 ≤ n ≤ 2·10⁵, 1 ≤ k ≤ m ≤ 2·10⁵, 1 ≤ s ≤ 10⁹) — number of days, total number and required number of gadgets, number of burles Nura has.

Second line contains n integers a_i (1 ≤ a_i ≤ 10⁶) — the cost of one dollar in burles oni-th day.

Third line contains n integers b_i (1 ≤ b_i ≤ 10⁶) — the cost of one pound in burles oni-th day.

Each of the next m lines contains two integerst_i, c_i (1 ≤ t_i ≤ 2, 1 ≤ c_i ≤ 10⁶) — type of the gadget and it's cost. For the gadgets of the first type cost is specified in dollars. For the gadgets of the second type cost is specified in pounds.

Output

If Nura can't buy k gadgets print the only line with the number-1.

Otherwise the first line should contain integer d — the minimum day index, when Nura will havek gadgets. On each of the next k lines print two integers q_i, d_i — the number of gadget and the day gadget should be bought. All valuesq_i should be different, but the valuesd_i can coincide (so Nura can buy several gadgets at one day). The days are numbered from1 to n.

In case there are multiple possible solutions, print any of them.

Sample test(s)

Input

5 4 2 21 2 3 2 13 2 1 2 31 12 11 22 2

Output

31 12 3

Input

4 3 2 20069 70 71 72104 105 106 1071 12 21 2

Output

-1

Input

4 3 1 1000000000900000 910000 940000 990000990000 999000 999900 9999901 876542 765431 65432

Output

-1

分析：英语太烂，理解题意好难。。。看了这篇题解才明白点击打开链接

题意：一个人手上有s卢布，他要在n天内买m样东西中的k样
样例第二行是多少卢布换一美刀，第三行是多少卢布换一英镑
然后接下来是m样东西，告诉你每样东西用什么货币结账（1是美刀，2是卢布），以及要多少那种外币
要求：最早完成买k样的天数，以及输出方案
输出第几天买完k样商品，然后k行
购买的商品的id，第几天购买的
输出任意答案就可以了
 一种商品只能购买一次，但是一天可以买多种商品 

思路：对于前i天购买东西，肯定是选择在1天内或者2天内买完，就是汇率最高的时候购买。

对于某一天，我们能用O(n)的复杂度来求出当前最低所需要的钱来买k件商品，这里面只要把商品按照购买方式分类，然后排序一遍，之后就很好维护了。

那么对于前i天，很明显这个有单调性，先求出前i天里面两种汇率最大的分别是哪两天，预处理一下，之后只要二分i，就能得到答案了

#include<bits/stdc++.h>using namespace std;#define INF 0x3f3f3f3fconst int maxn=200009;int a[maxn],b[maxn],t[maxn],c[maxn];vector<pair<long long, pair<int, int> > >items;int n,m,s,k;bool solve(int day){    items.clear();int mina=INF,minb=INF;int posa,posb;for (int i=1; i<=day; i++) {if (a[i]<mina) {mina=a[i];posa=i;}if (b[i]<minb) {minb=b[i];posb=i;}}for (int i=1; i<=m; i++) {if (t[i] == 1) items.push_back(make_pair(1ll*mina*c[i], make_pair(i, posa)));else items.push_back(make_pair(1ll*minb*c[i], make_pair(i, posb)));}long long sum=0;sort(items.begin(), items.end());for (int i=0; i<k; i++)sum += items[i].first;return s >= sum;}int main(){    scanf("%d%d%d%d",&n,&m,&k,&s);    for(int i=1;i<=n;i++)scanf("%d",&a[i]);    for(int i=1;i<=n;i++)scanf("%d",&b[i]);    for(int i=1;i<=m;i++)scanf("%d%d",&t[i],&c[i]);    int l=0,r=n+1;    while(l+1<r){        int mid=(l+r)/2;        if(solve(mid))r=mid;        else l=mid;    }    if (l == n) {        printf("-1\n");        return 0;}printf("%d\n", r);solve(r);for (int i=0;i<k; i++) {printf("%d %d\n", items[i].second.first, items[i].second.second);}    return 0;}