PAT (Advanced Level) 1033. To Fill or Not to Fill (25)

贪心。注意x=0处没有加油站的情况。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

struct X
{
    double cost, x, v;
    int id;
}s[ + ];
double C, D, P;
double len;
int n;

bool f(double a, double b)
{
    if (fabs(a - b)<1e-) return ;
    return ;
}

struct Y
{
    int id;
    double cost, x, v;
    Y(int ID, double COST, double X,double V)
    {
        id = ID;
        cost = COST;
        x = X;
        v = V;
    }
    bool operator < (const Y &a) const {
        if (f(cost, a.cost)) return x>a.x;
        return cost>a.cost;
    }
};

bool cmp(const X&a, const X&b) { return a.x<b.x; }

bool FAIL()
{
    len = ;
    if (s[].x > ) return ;
    for (int i = ; i < n; i++)
    {
        len = s[i].x + P*C;
        if (len < s[i + ].x) return ;
    }
    return ;
}

int main()
{
    scanf("%lf%lf%lf%d", &C, &D, &P, &n);
    for (int i = ; i <= n; i++)
    {
        scanf("%lf%lf", &s[i].cost, &s[i].x);
        s[i].v = ;
    }
    sort(s + , s +  + n, cmp);
    if (s[n].x<D)  n++, s[n].x = D;
    for (int i = ; i <= n; i++) s[i].id = i;

    if (FAIL()) printf("The maximum travel distance = %.2lf\n", len);
    else
    {
        double sum = , ans = ;
        int p = ;

        priority_queue<Y>Q;
        Q.push(Y(, s[].cost, s[].x, ));

        for (int i = ; i <= n; i++)
        {
            double d = s[i].x - s[i - ].x;
            double need = d / P;

            while ()
            {
                if (f(need, )) break;
                while ()
                {
                    Y head = Q.top(); Q.pop();
                    if (head.id < p) continue;
                    else if (f(s[head.id].v, C)) continue;
                    else
                    {
                        p = head.id;
                        double h = min(need, C - s[p].v);
                        need = need - h;
                        sum = sum + h;
                        s[p].v = sum - s[p].x / P;
                        ans = ans + s[p].cost*h;
                        Q.push(head);
                        break;
                    }
                }
            }

            Q.push(Y(i, s[i].cost, s[i].x, ));
        }
        printf("%.2lf\n", ans);
    }
    return ;
}