标签：poj 优先权队列
Expedition
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 8053 |
|
Accepted: 2359 |
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
4
4 4
5 2
11 5
15 10
25 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
题目链接：http://poj.org/problem?id=2431
题目大意：有n个加油站，一辆卡车开始离目的地的距离为l且有p升油，每个加油站有两个值，离目的地的距离和储油量，现在问卡车从起点到终点最少要加几次油，每次加油认为是将加油站油全加完
题目分析：首先要注意加油站的距离是相对目的地的，我们要将其转为相对于卡车的距离，即用l减，我们通过计算卡车从当前点到下一个加油站需要的油量，若不够则在之前的加油站中补，而且为了使加油次数最小，显然选择储油量最大的加油站先加，这样就选定了本题优先权队列的数据结构，每到一个加油站，将其入队列，补油时直接从队列里找，队列为空时说明不可达，注意这题的输入不一定按距离顺序，因此需要排个序
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
int const MAX = 1e4 + 5;
struct Stop
{
int dis, fuel;
}s[MAX];
bool cmp(Stop a, Stop b)
{
return a.dis < b.dis;
}
int main()
{
int n, l, p;
scanf("%d", &n);
for(int i = 0; i < n; i++)
scanf("%d %d", &s[i].dis, &s[i].fuel);
scanf("%d %d", &l, &p);
for(int i = 0; i < n; i++)
s[i].dis = l - s[i].dis;
s[n].fuel = 0; //将终点加进去
s[n++].dis = l;
sort(s, s + n, cmp);
priority_queue <int> q;
int ans = 0, pos = 0, num = p;
for(int i = 0; i < n; i++)
{
int d = s[i].dis - pos;
while(num < d)
{
if(q.empty())
{
printf("-1\n");
return 0;
}
ans ++; //加油
num += q.top();
q.pop();
}
num -= d;
q.push(s[i].fuel);
pos = s[i].dis;
}
printf("%d\n", ans);
}
POJ 2431 Expedition (STL 优先权队列)
标签：poj 优先权队列
原文地址：http://blog.csdn.net/tc_to_top/article/details/44840679
