zoj 3699 单调队列

最新推荐文章于 2019-03-29 11:19:48 发布

原创最新推荐文章于 2019-03-29 11:19:48 发布 · 711 阅读

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本文介绍了一个算法问题，旨在帮助达喀尔拉力赛车手通过最优加油策略来最小化比赛中的燃油成本。考虑到不同路段的里程数、车辆消耗及加油站油价的变化，文章提供了一段C++代码实现该策略。

Dakar Rally

Time Limit: 2 Seconds Memory Limit: 65536 KB

Description

The Dakar Rally is an annual Dakar Series rally raid type of off-road race, organized by the Amaury Sport Organization. The off-road endurance race consists of a series of routes. In different routes, the competitors cross dunes, mud, camel grass, rocks, erg and so on.

Because of the various circumstances, the mileages consume of the car and the prices of gas vary from each other. Please help the competitors to minimize their payment on gas.

Assume that the car begins with an empty tank and each gas station has infinite gas. The racers need to finish all the routes in order as the test case descripts.

Input

There are multiple test cases. The first line of input contains an integer T (T ≤ 50) indicating the number of test cases. Then T test cases follow.

The first line of each case contains two integers: n -- amount of routes in the race; capacity -- the capacity of the tank.

The following n lines contain three integers each: mileage_i -- the mileage of the i^th route; consume_i -- the mileage consume of the car in the i^th route , which means the number of gas unit the car consumes in 1 mile; price_i -- the price of unit gas in the gas station which locates at the beginning of the i^th route.

All integers are positive and no more than 10⁵.

Output

For each test case, print the minimal cost to finish all of the n routes. If it's impossible, print "Impossible" (without the quotes).

Sample Input

Sample Output

550
Impossible


#include <iostream>
#include <cstring>
#include <cstdio>
#include <deque>

using namespace std;

#define maxn 100005

struct Data
{
    int m, c, p;

}a[maxn];

int tot[maxn];
int n, cap;

int main()
{
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d %d", &n, &cap);

        for(int i = 0; i < n; i++)
            scanf("%d %d %d", &a[i].m, &a[i].c, &a[i].p);

        deque<int> que;
        memset(tot, 0, sizeof(tot));
        long long ans = 0;
        for(int i = 0; i < n; i++){
            while(!que.empty() && a[que.back()].p >= a[i].p) que.pop_back();
            que.push_back(i);

            long long need = (long long)a[i].m*a[i].c;
            if(need > cap){
                ans = -1;
                break;
            }

            int pos;
            int cur_cap = 0, inc;
            while(cur_cap < need && !que.empty()){
                pos = que.front();
                inc = min(need-cur_cap, (long long)cap-tot[pos]);
                cur_cap += inc;
                ans += (long long)inc*a[pos].p;
                for(int j = pos; j <= i; j++)
                    tot[j] += inc;
                if(tot[pos] == cap)
                    que.pop_front();
            }

            tot[i] = need;

        }

        if(ans == -1)
            printf("Impossible\n");
        else
            cout << ans << endl;
    }
    return 0;
}