poj2431 Expedition （优先队列） 挑战程序设计竞赛

题目链接：http://poj.org/problem?id=2431

Expedition

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9148 Accepted: 2673

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.

这是挑战程序设计竞赛上的关于 堆 以及 优先队列运用的例题，书上给出的思路非常巧妙

因为求得是最优解，需要尽可能少的加油站，所以我们每次希望去加油的时候 加最大的那个，因而将加油站push进priority_queue(堆结构，默认每次弹出最大值)

从起点开始跑，把路程通过加油站分成几部分，每一部分只要通过加油能到达加油站即可，，因而，把终点也视为加油站 距离L，加油量 0；

#include<cstdio> //输入量较大，所以用c风格来输入输出， ac时间 16ms ， 全部换成c++的风格则需要 313ms#include<queue>#include<algorithm>using namespace std;struct node{int a1,b1;};int cmp(const node& a,const node& b){return a.a1<b.a1;}int main(){priority_queue<int> pq;int n;node no[10010];while(scanf("%d",&n)!=EOF){for(int i=0;i<n;i++){scanf("%d%d",&no[i].a1,&no[i].b1);}int L,P,cnt=0,sign=0;scanf("%d%d",&L,&P);for(int i=0;i<n;i++){  //题目要求给的a1是当前与终点的距离，我们转换为与起点的距离no[i].a1=L-no[i].a1;}sort(no,no+n,cmp);int index=0;//pq.push(P);int pos=0;no[n].a1=L;  //将终点设为加油站no[n].b1=0;n++;for(int i=0;i<n;i++){ int d=no[i].a1-pos; //当前需要走的距离 while(P<d){   if(pq.empty()){ //油箱为空 puts("-1"); sign=1; break; } else{ P+=pq.top(); pq.pop(); cnt++; } } P-=d; pos=no[i].a1; pq.push(no[i].b1);}if(!sign){printf("%d\n",cnt);}}return 0;}

下面是另一种方法

#include <cstdio>#include <algorithm>#include <queue>using namespace std;struct node{int a1,b1;}no[10010];int cmp(const node&a, const node&b){return a.a1<b.a1;}int main(){int t;priority_queue<int>pq;while(scanf("%d",&t)!=EOF){while(!pq.empty())pq.pop();for(int i=0;i<t;i++){scanf("%d%d",&no[i].a1,&no[i].b1);}int l,p;scanf("%d%d",&l,&p);for(int i=0;i<t;i++){no[i].a1=l-no[i].a1;}sort(no,no+t,cmp);int cur=p;//当前位置int i=0;int cnt=0;while(cur<l){ //如果还没到终点 while(no[i].a1<=cur&&i<t){ //能加油就尽量加pq.push(no[i].b1);i++;}if(pq.empty()) break;cur+=pq.top(); 当前位置往前移动 （加上当前最大的加油站里的油）pq.pop();cnt++;}if(cur>=l) //当最远跑的距离超过终点printf("%d\n",cnt);elseprintf("-1\n");}return 0;}