POJ 2431 Expedition 优先队列

Expedition

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9775 Accepted: 2837

Description

A group of cows grabbed a truck and ventured on an expedition deep into the jungle. Being rather poor drivers, the cows unfortunately managed to run over a rock and puncture the truck's fuel tank. The truck now leaks one unit of fuel every unit of distance it travels. 

To repair the truck, the cows need to drive to the nearest town (no more than 1,000,000 units distant) down a long, winding road. On this road, between the town and the current location of the truck, there are N (1 <= N <= 10,000) fuel stops where the cows can stop to acquire additional fuel (1..100 units at each stop). 

The jungle is a dangerous place for humans and is especially dangerous for cows. Therefore, the cows want to make the minimum possible number of stops for fuel on the way to the town. Fortunately, the capacity of the fuel tank on their truck is so large that there is effectively no limit to the amount of fuel it can hold. The truck is currently L units away from the town and has P units of fuel (1 <= P <= 1,000,000). 

Determine the minimum number of stops needed to reach the town, or if the cows cannot reach the town at all. 

Input

* Line 1: A single integer, N 

* Lines 2..N+1: Each line contains two space-separated integers describing a fuel stop: The first integer is the distance from the town to the stop; the second is the amount of fuel available at that stop. 

* Line N+2: Two space-separated integers, L and P

Output

* Line 1: A single integer giving the minimum number of fuel stops necessary to reach the town. If it is not possible to reach the town, output -1.

Sample Input

44 45 211 515 1025 10

Sample Output

2

Hint

INPUT DETAILS: 

The truck is 25 units away from the town; the truck has 10 units of fuel. Along the road, there are 4 fuel stops at distances 4, 5, 11, and 15 from the town (so these are initially at distances 21, 20, 14, and 10 from the truck). These fuel stops can supply up to 4, 2, 5, and 10 units of fuel, respectively. 

OUTPUT DETAILS: 

Drive 10 units, stop to acquire 10 more units of fuel, drive 4 more units, stop to acquire 5 more units of fuel, then drive to the town.

Source

USACO 2005 U S Open Gold

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  给你一辆车，初始状态下有p升汽油每行驶一公里消耗一升汽油。在前往距离为len的目的地上有n个加油站，给出加油站距离终点的距离和加油站可以提供的汽油ni升。假设汽车的油箱能够装下足够多的汽油，问汽车最多加多少次汽油能够到达目的地。

  起先汽车能跑p里路，假设有汽油的时候经过加油站先不加油，那么它最多能跑p里。为了能到目的地肯定在这之前经过的加油站加油，为了加油次数最少，就要选择此前经过的加油站中油数最多的那一站加油。

  我们用优先队列存储经过的加油站，表示能够在此站加油，当汽车没油的时候选择一个最大的加油，这样就可以得到答案。

ACcode：

#pragma warning(disable:4786)//使命名长度不受限制#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈#include <map>#include <set>#include <queue>#include <cmath>#include <stack>#include <cctype>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rds(x) scanf("%s",x)#define rdc(x) scanf("%c",&x)#define ll long long int//#define maxn 1e5+5#define mod 1000000007#define INF 0x3f3f3f3f //int 最大值#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define MT(x,i) memset(x,i,sizeof(x))#define PI  acos(-1.0)#define E  exp(1)const int maxn=1e5+5;using namespace std;int len,p,n,pos,tank,ans,d;struct N{    int a,b;    void in(){rd2(a,b);}    void geta(){a=len-a;}///给出的是距离终点的距离而我们需要的是距离起点的距离}my[maxn];bool cmp(N x,N y){return x.a<y.a;}bool flag;int main(){    rd(n);MT(my,0);    FOR(i,0,n-1)my[i].in();    rd2(len,p);    FOR(i,0,n-1)my[i].geta();    ans=pos=0;    my[n].a=len;my[n].b=0;    tank=p;n++;    sort(my,my+n,cmp);    priority_queue<int> q;    FOR(i,0,n-1){        d=my[i].a-pos;        while(tank<d){            if(q.empty()){                puts("-1\n");                return 0;            }            tank+=q.top();            q.pop();            ans++;        }        tank-=d;        pos=my[i].a;        q.push(my[i].b);    }    printf("%d\n",ans);    return 0;}/*44 45 211 515 1025 10*/