Codeforces round #380 (Div. 2) C. Road to cinema binary

Question:

1. Give you N vehicles, K gas stations, S-route, within T seconds

2. There are two modes for all vehicles: 1 km/min and 1 km/2 min. The fuel consumption is 2/km and 1/km.

3. Each vehicle has a price and a tank capacity. When passing through a gas station, gasoline can be filled with blood instantly. Find the cheapest car.

Ideas:

1. Write a bool function judge to judge whether the vehicle spends the minimum time mint. If mint <= t, the system returns true. Use a gas station to group the routes, find the mints for each group, and accumulate them.

2. There are three cases of DIS in each group:

A. The tank capacity feul <DIS, which indicates that the slow speed is too big. Return false;

B. Tank Capacity fuel> = 2 * Dis, indicating that the tank can be reached directly at a high speed. The speed is 1 km/min, so mint + = DIS;

C. The tank capacity is between AB and use high speed as much as possible. High speed X km, low speed y km.

X + y = DIS, 2x + Y <= feul.

Obtain the value of X <= fuel-Dis, that is, the maximum value of X is fuel-dis. Mint + = x + 2 * Y.

3. The maximum tank volume in the vehicle maxcompute uses judge once. If not, output-1 directly. Otherwise, the minimum tank capacity (0, maxfuel) that can be reached on time is obtained.

4. Scan the car. If the capacity of the car is greater than the minimum capacity, Res = min (Res, car [I]. cost );

The Code is as follows:

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#include<iostream>#include<algorithm>#include<cstdio>using namespace std;const int maxn=200200;struct Car{    int cost, fuel;}car[maxn];int gas[maxn];bool judge(int fuel,int k,int t){    /*        x+y=dis        2x+y<=fuel        so x<=fuel-dis        x represent the distance with high speed, y ...    */    int i,j;    int mint=0;    for(i=1;i<=k+1;i++)    {        int dis=gas[i]-gas[i-1];        if(fuel<dis)            return false;        else if(fuel>=2*dis)            mint+=dis;        else        {            int x=fuel-dis;            int y=dis-x;            mint+=x+2*y;        }    }    return mint<=t;}int main(){    int n,k,s,t,i,j;    scanf("%d%d%d%d",&n,&k,&s,&t);    int maxfuel=-1;    for(i=1;i<=n;i++)        scanf("%d%d",&car[i].cost,&car[i].fuel),        maxfuel=max(maxfuel,car[i].fuel);    for(i=1;i<=k;i++)        scanf("%d",&gas[i]);    sort(gas+1,gas+k+1);    gas[0]=0;    gas[k+1]=s;    if(!judge(maxfuel,k,t))    {        printf("-1\n");        return 0;    }    int low,high;    low=0,high=maxfuel;    while(low    {        int mid=low+high >> 1;        if(judge(mid,k,t))            high=mid;        else            low=mid+1;    }    int res=2e9+5;    for(i=1;i<=n;i++)        if(car[i].fuel>=low)            res=min(res,car[i].cost);    printf("%d\n",res);}

Codeforces round #380 (Div. 2) C. Road to cinema binary