【26.83%】【Codeforces Round #380C】Road to Cinema

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Vasya is currently at a car rental service, and he wants to reach cinema. The film he has bought a ticket for starts in t minutes. There is a straight road of length s from the service to the cinema. Let’s introduce a coordinate system so that the car rental service is at the point 0, and the cinema is at the point s.

There are k gas stations along the road, and at each of them you can fill a car with any amount of fuel for free! Consider that this operation doesn’t take any time, i.e. is carried out instantly.

There are n cars in the rental service, i-th of them is characterized with two integers ci and vi — the price of this car rent and the capacity of its fuel tank in liters. It’s not allowed to fuel a car with more fuel than its tank capacity vi. All cars are completely fueled at the car rental service.

Each of the cars can be driven in one of two speed modes: normal or accelerated. In the normal mode a car covers 1 kilometer in 2 minutes, and consumes 1 liter of fuel. In the accelerated mode a car covers 1 kilometer in 1 minutes, but consumes 2 liters of fuel. The driving mode can be changed at any moment and any number of times.

Your task is to choose a car with minimum price such that Vasya can reach the cinema before the show starts, i.e. not later than in t minutes. Assume that all cars are completely fueled initially.

Input

The first line contains four positive integers n, k, s and t (1 ≤ n ≤ 2·105, 1 ≤ k ≤ 2·105, 2 ≤ s ≤ 109, 1 ≤ t ≤ 2·109) — the number of cars at the car rental service, the number of gas stations along the road, the length of the road and the time in which the film starts.

Each of the next n lines contains two positive integers ci and vi (1 ≤ ci, vi ≤ 109) — the price of the i-th car and its fuel tank capacity.

The next line contains k distinct integers g1, g2, …, gk (1 ≤ gi ≤ s - 1) — the positions of the gas stations on the road in arbitrary order.

Output

Print the minimum rent price of an appropriate car, i.e. such car that Vasya will be able to reach the cinema before the film starts (not later than in t minutes). If there is no appropriate car, print -1.

Examples

input

3 1 8 10

10 8

5 7

11 9

3

output

10

input

2 2 10 18

10 4

20 6

5 3

output

20

Note

In the first sample, Vasya can reach the cinema in time using the first or the third cars, but it would be cheaper to choose the first one. Its price is equal to 10, and the capacity of its fuel tank is 8. Then Vasya can drive to the first gas station in the accelerated mode in 3 minutes, spending 6 liters of fuel. After that he can full the tank and cover 2 kilometers in the normal mode in 4 minutes, spending 2 liters of fuel. Finally, he drives in the accelerated mode covering the remaining 3 kilometers in 3 minutes and spending 6 liters of fuel.

【题目链接】:http://codeforces.com/contest/738/problem/C

【题解】



把每个gas station按照位置升序排;

把车按照v[i]升序排(c[i]不用管);

l = 1,r = n;

m = (l+r)>>1;

这个m作为当前枚举的车是什么;

注意到每次到车站之后油都能变满(哪个傻子不装满?免费的!)

它的容量作为从两个gas站之间的花费上限;

while(l<=r)
{
    ...
    if (ok(m))
        ans = m,r = m-1;
    else
        l = m+1;
}

那么ok函数要怎么写?

    bool ok(int w)
    {
        LL temp =0;
        for (i = 1->k)
        {
            int ds = p[i]-p[i-1];
            if (ds*2<=w)//使用加速模式花费汽油是距离的两倍
                temp+=ds;//所用的时间就是距离;
            else//不能全部使用加速模式
            {//那就设低速模式x分钟,加速模式y分钟
/*
    则有
    0.5*x+y=ds ···①距离要恰好为两个站之间的距离
    0.5x+2*y<=w ···②花费的燃油要小于等于油箱容量
    显然要让y最大,这样时间最短;
    而联立①②式可得
    y<=w-ds;
    所以y的最大值为w-ds;
    相应的x = 2*(ds-y);
*/
                if (w-ds>0)//如果能够有一段为加速模式
                {
                    y = w-ds;
                    x = 2*(ds-y);
                    cost = y*2+x/2;
                    temp+=(x+y);//时间就是x+y;
                 }
                 else//只能全程为低速模式尝试一下
                 {
                    if (ds<=w)
                        temp+=2*ds;//时间为两倍距离
                    else
                        return false;
                 }
            }
            if (temp>t)//时间大于t了就返回false;
                return false;
        }
        return true;
    }

这样我们就把油箱容量符合要求的最小油箱容量的车的下标ans搞出来了；

之后枚举1->n，找到那些油箱容量大于等于这个ans车的容量的车；在它们之间找价格最低的；



【完整代码】

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int MAXNK = 2e5+10;
const LL INF = 1e18;
struct abc
{
    int c,v;
};
abc a[MAXNK];
int p[MAXNK];
int n,k,s;
LL t;
LL fi = INF;
bool cmp1(abc a,abc b)
{
    return a.v < b.v;
}
bool ok(int w)
{
    LL temp = 0;
    for (int i = 1;i <= k;i++)
        {
            int cost = (p[i]-p[i-1])*2;
            if (cost<=w)//直接全速前进
                temp+=p[i]-p[i-1];
            else
            {//枚举在p[i-1]..p[i],前面用低速后面用高速;
                int ds = p[i]-p[i-1];
                int x,y;
                if (w-ds>0)
                {
                    y = w-ds;
                    x = 2*(ds-y);
                    cost = y*2+x/2;
                    temp+=(x+y);
                    /*
                    if (w==8)
                    {
                        puts("233");
                        cout << "ds==" << ds<<endl;
                        cout << "temp=="<<temp<<endl;
                        cout << "w-ds=="<<w-ds<<endl;
                        cout <<"cost=="<<cost<<endl;
                        cout << y<<endl;
                        cout << x<<endl;
                    }
                    */
                }
                else
                {
                    if (ds<=w)
                        temp+=2*ds;
                    else
                        return false;
                }
            }
            if (temp > t)
                return false;
        }
    return true;
}
int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    cin >> n >> k >>s >>t;
    for (int i = 1;i <= n;i++)
        scanf("%d%d",&a[i].c,&a[i].v);
    p[0] = 0;
    for (int i = 1;i <= k;i++)
        scanf("%d",&p[i]);
    sort(a+1,a+1+n,cmp1);
    sort(p+1,p+1+k);
    k++;
    p[k] = s;
    /*
    for (int i = 1;i <= k;i++)
        cout << p[i] << " ";
    cout << endl;
    */
    int l = 1,r = n,ans = -1;
    while (l <= r)
    {
        int m = (l+r)>>1;
        if (ok(a[m].v))
        {
            ans = m;
            r = m-1;
        }
        else
            l = m+1;
    }
    if (ans == -1)
    {
        puts("-1");
        return 0;
    }
    //cout << ans << endl;
    for (int i = 1;i <= n;i++)
    {
        if (a[i].v >= a[ans].v && a[i].c < fi)
            fi = a[i].c;
    }
    cout << fi << endl;
         //wujie -1
    return 0;
}