POJ - 2431 Expedition(贪心+优先队列)

题意：已知某车距离城镇为L，油量为P，油箱无穷大，已知有n个加油量为x的加油站，问，车到城镇最少加几次油。若不能到达城镇，则输出-1。

分析：

1、贪心，先将加油站按照离城镇由远及近排序。

2、卡车只要油够，就不断往前走，若当前油量不足以到达终点（或下一个加油站），则在之前经过的加油站里选择提供油量最大的加油站，加够了再继续走。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 10000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int N, L, P;
struct Node{
    int d, fuel;
    Node(){}
    void read(){
        scanf("%d%d", &d, &fuel);
    }
    void Set(int dd, int ff){
        d = dd;
        fuel = ff;
    }
    bool operator < (const Node& rhs)const{
        return d > rhs.d;
    }
}num[MAXN];
priority_queue<int> q;
int solve(){
    int ans = 0;
    for(int i = 1; i <= N + 1; ++i){
        int tmp = num[i - 1].d - num[i].d;
        if(P - tmp >= 0) P -= tmp;
        else{
            while(!q.empty() && P - tmp < 0){
                P += q.top();
                q.pop();
                ++ans;
            }
            if(P - tmp < 0) return -1;
            P -= tmp;
        }
        q.push(num[i].fuel);
    }
    return ans;
}
int main(){
    scanf("%d", &N);
    for(int i = 1; i <= N; ++i){
        num[i].read();
    }
    scanf("%d%d", &L, &P);
    sort(num + 1, num + 1 + N);
    num[0].Set(L, 0);
    num[N + 1].Set(0, 0);
    printf("%d\n", solve());
    return 0;
}