LuoguP1016 旅行家的预算 （贪心）

胡一个错误代码都能有75pts

忘了怎么手写deque其实是懒

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); (a) <= (c); ++(a))
#define nR(a,b,c) for(register int a = (b); (a) >= (c); --(a))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))

//#define ON_DEBUGG

#ifdef ON_DEBUGG

#define D_e_Line printf("\n----------\n")
#define D_e(x) cout << (#x) << " : " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt", "r", stdin)

#else

#define D_e_Line ;
#define D_e(x) ;
#define Pause() ;
#define FileOpen() ;

#endif
using namespace std;
struct ios{
	template<typename ATP>inline ios& operator >> (ATP &x){
		x = 0; int f = 1; char ch;
		for(ch = getchar(); ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;
		while(ch >= '0' && ch <= '9') x = x * 10 + (ch ^ '0'), ch = getchar();
		x *= f;
		return *this;
	}
}io;

template<typename ATP>inline ATP max(ATP &a, ATP &b){
	return a > b ? a : b;
}
#include <deque>

struct Petrol {
    double cost, volume;
};

deque<Petrol> dq;
double dis[17], price[17];

int main() {
FileOpen();

	double totDis, fuelMax, dx;
	int n;
    scanf("%lf%lf%lf%lf%d", &totDis, &fuelMax, &dx, &price[0], &n);
    R(i,1,n){
        scanf("%lf%lf", &dis[i], &price[i]);
        if(dis[i] - dis[i - 1] > fuelMax * dx){
            printf("No Solution\n");
            return 0;
        }
    }

    dis[n+1] = totDis;
    double fuelNow = fuelMax;
    dq.push_back((Petrol){price[0], fuelMax});
    double ans = price[0] * fuelMax;
    R(i,1,i + 1){
        double costFuel = (dis[i] - dis[i - 1]) / dx;
        while(!dq.empty() && costFuel > 0){
            Petrol x = dq.front(); dq.pop_front();
            if(x.volume > costFuel){
                fuelNow -= costFuel;
                dq.push_front((Petrol){x.cost, x.volume - costFuel});
                break;
            }
            fuelNow -= x.volume;
			costFuel -= x.volume;
        }

        if(i == n + 1){
            while(!dq.empty()) {
                ans -= dq.front().cost * dq.front().volume;
                dq.pop_front();
            }
            break;
        }

        while(!dq.empty() && dq.back().cost > price[i]){
            ans -= dq.back().cost * dq.back().volume;
            fuelNow -= dq.back().volume;
            dq.pop_back();
        }

        ans += (fuelMax - fuelNow) * price[i];
        dq.push_back((Petrol){price[i], fuelMax - fuelNow});
        fuelNow = fuelMax;
    }

    printf("%.2lf\n", ans);
    return 0;
}