Challenges Programming Contest: Expeition

Subject to the effect

Problem-solving ideas

• After the gas station as a chance to refuel, pile into the big top, when no oil when refueling can be removed from the big top of the heap up to increase opportunities for oil.
  

Code

#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;

const int MAXN = 10005;
const int inf = 1000000;
pair<int, int> stops[MAXN];
priority_queue<int, vector<int> > PQ;
int main()
{
    int N;
    int L;
    int left_tank;
    cin >> N;
    for(int i=0; i<N; i++)
        cin >> stops[i].first >> stops[i].second;
    cin >> L >> left_tank;
    for(int i=0; i<N; i++)
        stops[i].first = L - stops[i].first;
    sort(stops, stops+N);
    int now_pos = 0;
    int last_stop = 0; //指向上一个还没有纳入机会集合的第一个加油站， 初始为0
    int now_stop; //指向前进一次后最后一个能纳入机会集合的加油站
    int ans = 0;
    while(now_pos + left_tank < L) //如果>=，则当前油量足够到达终点
    {
        now_pos += left_tank;
        now_stop = upper_bound(stops, stops+N, make_pair(now_pos, inf)) - stops - 1;
        for(int i=last_stop; i<=now_stop; i++)
            PQ.push(stops[i].second);
        last_stop = now_stop + 1;
        if(PQ.empty())
        {
            cout << -1 << endl;
            return 0;
        }
        left_tank = PQ.top();
        ans++;
        PQ.pop();
    }
    cout << ans << endl;
    return 0;
}

Knowledge Point

• upper_boundAnd pairthe combined use of: (1) press the firstsort, then secondsort, so here looking for a last chance gas station can be incorporated into the set, will secondset up in order infto ensure the return of the elements firstmust be greater than now_pos. (2) make_pairdirect generation of a pair.