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
題目鏈接：http://poj.org/problem?id=2431
題目大意：有n個加油站，一輛卡車開始離目的地的距離為l且有p升油，每個加油站有兩個值，離目的地的距離和儲油量，現在問卡車從起點到終點最少要加幾次油，每次加油認為是將加油站油全加完
題目分析：首先要注意加油站的距離是相對目的地的，我們要將其轉為相對於卡車的距離，即用l減，我們通過計算卡車從當前點到下一個加油站需要的油量，若不夠則在之前的加油站中補，而且為了使加油次數最小，顯然選擇儲油量最大的加油站先加，這樣就選定了本題優先權隊列的數據結構，每到一個加油站，將其入隊列，補油時直接從隊列裡找，隊列為空時說明不可達，注意這題的輸入不一定按距離順序，因此需要排個序
#include
#include
#include
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; //將終點加進去
s[n++].dis = l;
sort(s, s + n, cmp);
priority_queue 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);
}
