POJ - 2431 - Expedition（贪心 + 优先队列）

题意：给定一条 L 长的路（1 <= L <= 10^6），途中有 n 个加油站（1 <= n <= 10^4），其中第 i 个加油站在距离终点Ai（1 <= Ai <= L）距离的地方，可以加Bi（1 <= Bi <= 100）单位油，设一单位距离耗一单位油，起点时有P（1 <= P <= 10^6）升油，求到达终点最少需要加多少次油，若无法到达终点则输出-1。

可以将此过程看作：每次尽可能走，直到消耗完目前所有的油，途中记录所有经过的加油站，然后在没油的时候取能加油量最大的加油站加油即可。

代码如下：

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define pr(x) cout << #x << " : " << x << "   "
#define prln(x) cout << #x << " : " << x << endl
#define Min(a, b) a < b ? a : b
#define Max(a, b) a < b ? b : a
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const ll MOD = 1e9 + 7;
using namespace std;

#define NDEBUG
#include<cassert>
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;

struct Node{
    int cur, fuel;
    bool operator < (const Node &rhs) const{
        return fuel < rhs.fuel;
    }
}p[MAXT];

bool cmp(const Node &a, const Node &b){
    return a.cur < b.cur || (a.cur == b.cur && a.fuel > b.fuel);
}

int n, length, y;

int main(){
    while(scanf("%d", &n) == 1){
        for(int i = 0; i < n; ++i)  scanf("%d%d", &p[i].cur, &p[i].fuel);
        scanf("%d%d", &length, &y);
        for(int i = 0; i < n; ++i)  p[i].cur = length - p[i].cur;
        p[n] = (Node){length, 0};
        sort(p, p + n + 1, cmp);
        priority_queue<Node> q;
        bool flag = true;
        int lur = 0, fuels = y, ans = 0;
        for(int i = 0; i <= n; ++i){
            if(p[i].cur - lur <= fuels){
                fuels -= p[i].cur - lur;
                lur = p[i].cur;
                q.push(p[i]);
            }
            else{
                while(!q.empty() && p[i].cur - lur > fuels){
                    const Node &u = q.top();
                    fuels += u.fuel;
                    q.pop();
                    ++ans;
                }
                if(p[i].cur - lur <= fuels){
                    fuels -= p[i].cur - lur;
                    lur = p[i].cur;
                    q.push(p[i]);
                }
                else{
                    flag = false;
                    break;
                }
            }
        }
        printf("%d\n", flag ? ans : -1);
    }
    return 0;
}